Since the enthalpy of elements in their natural state is taken to be zero, the heat of formation `(Delta_(f)H)` of compounds

Since the enthalpy of elements in their natural state is taken to be zero, the heat of formation `(Delta_(f)H)` of compounds
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Since the enthalpy of elements in their natural state is taken to be zero, the heat of formation `( ... C. is zero D. May be positive or negative
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`Delta_(f)^(@)` for `CO_(2(g)), CO_((g))` and `H_(2)O_((g))` are `-393.5,-110.5` and `-241.8kJ mol^(-1)` respectively. The standard enthalpy change `(

Description : `Delta_(f)^(@)` for `CO_(2(g)), CO_((g))` and `H_(2)O_((g))` are `-393.5,-110.5` and `-241.8kJ mol^(-1)` respectively. The standard enthalpy change `(

Last Answer : `Delta_(f)^(@)` for `CO_(2(g)), CO_((g))` and `H_(2)O_((g))` are `-393.5,-110.5` and `-241.8kJ mol^(-1 ... A. `524.1` B. `41.2` C. `-262.5` D. `-41.2`

From the elements: Cl, Br, F, O, Al, C, Li, Cs and Xe; choose the following: (a) The element with highest negative electron gain enthalpy. (b) The element with lowest ionization enthalpy. (c) The element with smallest ionic radius. (d) The element with six electrons in the valence shell. (e) The element which is a liquid at room temperature. (f)The element which belongs to zero group. (g) The elements which forms largest number of compounds.

Description : From the elements: Cl, Br, F, O, Al, C, Li, Cs and Xe; choose the following: (a) The element with highest negative electron gain enthalpy. (b) The element with lowest ionization enthalpy. ( ... . (f)The element which belongs to zero group. (g) The elements which forms largest number of compounds.

Last Answer : Ans. (a). The element chlorine (Cl) has the highest negative electron gain enthalpy. (b). The element cesium (Cs) has lowest ionization enthalpy. (c). The element fluorine F has lowest atomic radius. (d ... zero group (or group 18). (g). The element carbon (C) forms the largest number of compounds.

Out of carbon(diamond)and carbon (graphite) whose enthalpy of formation is taken as zero?

Description : Out of carbon(diamond)and carbon (graphite) whose enthalpy of formation is taken as zero?

Last Answer : Ans. Enthalpy of formation of carbon in the form of graphite is taken as zero because it is a more commonly found in stable form of carbon.

For the given reaction , `A(g) rarr 2B(g) , Delta_(r)H= 30 kJ//"mole" Delta_(r)S = 150 J//mol` at `300 K` If `C_(P,A) = 20 J//K mol` and `C_(P,B) = 20

Description : For the given reaction , `A(g) rarr 2B(g) , Delta_(r)H= 30 kJ//"mole" Delta_(r)S = 150 J//mol` at `300 K` If `C_(P,A) = 20 J//K mol` and `C_(P,B) = 20

Last Answer : For the given reaction , `A(g) rarr 2B(g) , Delta_(r)H= 30 kJ//"mole" Delta_(r)S = 150 J//mol` at ... G` is negative D. At `T = 200, Delta G` is zero

Give enthalpy (H) of all elements in their standard state.

Description : Give enthalpy (H) of all elements in their standard state.

Last Answer : Ans. In standard state enthalpies of all elements is zero.

Enthalpy 'H' is defined as (A) H = E - PV (B) H = F - TS (C) H - E = PV (D) None of these

Description : Enthalpy 'H' is defined as (A) H = E - PV (B) H = F - TS (C) H - E = PV (D) None of these

Last Answer : (C) H - E = PV

Assertion: The enthalpy of formation of gaseous oxygen molecules at `298K` and under1 atm is zero. Reason: The entropy of formation of gaseous oxygen

Description : Assertion: The enthalpy of formation of gaseous oxygen molecules at `298K` and under1 atm is zero. Reason: The entropy of formation of gaseous oxygen

Last Answer : Assertion: The enthalpy of formation of gaseous oxygen molecules at `298K` and under1 atm is zero. ... D. If both Assertion & Reason are false.

Standard enthalpy of formation is zero for.

Description : Standard enthalpy of formation is zero for.

Last Answer : Standard enthalpy of formation is zero for. A. `C_("diamond")` B. `Br_((g))` C. `C_("graphite")` D. `O_(3(g))`

Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat of combustion when one mole of graphite bu

Description : Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat of combustion when one mole of graphite bu

Last Answer : Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat ... B. `-284` KJ C. `-394` KJ D. `-504` KJ

The enthalpy change when ammonia gas is dissolved in water is called the heat of (A) Solution (B) Formation (C) Dilution (D) Combustion

Description : The enthalpy change when ammonia gas is dissolved in water is called the heat of (A) Solution (B) Formation (C) Dilution (D) Combustion

Last Answer : (A) Solution

Given that standard heat enthalpy of `CH_(4), C_(2)H_(4)` and `C_(3)H_(8)` are -17.9, 12.5, -24.8 Kcal/mol. The `Delta H` for `CH_(4)+C_(2)Hrarr C_(3)

Description : Given that standard heat enthalpy of `CH_(4), C_(2)H_(4)` and `C_(3)H_(8)` are -17.9, 12.5, -24.8 Kcal/mol. The `Delta H` for `CH_(4)+C_(2)Hrarr C_(3)

Last Answer : Given that standard heat enthalpy of `CH_(4), C_(2)H_(4)` and `C_(3)H_(8)` are -17.9, 12.5, -24.8 Kcal ... . `-30.2` Kcal C. 55.2 Kcal D. `-19.4` Kcal

In the reaction, C + O2 → CO2; ΔH = - 94 kcal. What is the heat content (enthalpy) of O2? (A) -94 kcal (B) > -94 kcal (C) < - 94 kcal (D) Zero

Description : In the reaction, C + O2 → CO2; ΔH = - 94 kcal. What is the heat content (enthalpy) of O2? (A) -94 kcal (B) > -94 kcal (C) < - 94 kcal (D) Zero

Last Answer : (D) Zero

It is a common observation that naturally occurring processes have a particular direction associated with them. For instance, heat flows from a hot body to a colder body. The property of a system that is used to quantitatively describe this natural direction of events is called w) energy x) entropy y) enthalpy z) equilibrium

Description : It is a common observation that naturally occurring processes have a particular direction associated with them. For instance, heat flows from a hot body to a colder body. The property of a system ... this natural direction of events is called w) energy x) entropy y) enthalpy z) equilibrium 

Last Answer : ANSWER: X -- ENTROPY

The magnitude of crystal field stabilisation energy (CFSE of `Delta_(1)`) in tetrahedral complexes is considerably less than that in the octahderal fi

Description : The magnitude of crystal field stabilisation energy (CFSE of `Delta_(1)`) in tetrahedral complexes is considerably less than that in the octahderal fi

Last Answer : The magnitude of crystal field stabilisation energy (CFSE of `Delta_(1)`) in tetrahedral complexes is ... Both point (A) and (B) are wrong.

Among the following species the one which causes the highest `CFSE, Delta_(0)` as a ligands is :

Description : Among the following species the one which causes the highest `CFSE, Delta_(0)` as a ligands is :

Last Answer : Among the following species the one which causes the highest `CFSE, Delta_(0)` as a ligands is : A. `CN^(-)` B. `NH_(3)` C. `F^(-)` D. CO

Which complex of the following pairs has the larger value of `Delta_(0)` (i) `[Co(CN)_(6)]^(3-)` and `[Co(NH_(3))_(6)]^(3+)` (ii) `[Co(NH_(3))_(6)]^(3

Description : Which complex of the following pairs has the larger value of `Delta_(0)` (i) `[Co(CN)_(6)]^(3-)` and `[Co(NH_(3))_(6)]^(3+)` (ii) `[Co(NH_(3))_(6)]^(3

Last Answer : Which complex of the following pairs has the larger value of `Delta_(0)` (i) `[Co(CN)_(6)]^(3-)` and `[Co(NH_( ... (3))_(6)]^(3+) lt [CoF_(6)]^(3-)`

The magnitude of crystal field stabilisation energy (CFSE of `Delta_(1)`) in tetrahedral complexes is considerably less than that in the octahderal fi

Description : The magnitude of crystal field stabilisation energy (CFSE of `Delta_(1)`) in tetrahedral complexes is considerably less than that in the octahderal fi

Last Answer : The magnitude of crystal field stabilisation energy (CFSE of `Delta_(1)`) in tetrahedral complexes is ... Both point (A) and (B) are wrong.

The crystl field splitting energy for octahedral complex `(Delta_(0))` and that for tetrahedral complex `(Delta_(1))` rae related as :

Description : The crystl field splitting energy for octahedral complex `(Delta_(0))` and that for tetrahedral complex `(Delta_(1))` rae related as :

Last Answer : The crystl field splitting energy for octahedral complex `(Delta_(0))` and that for tetrahedral complex `(Delta_(1) ... ` D. `Deltat=(9)/(4)Delta_(0)`

The value of `Delta_(0) " for " [Ti(H_(2)O)_(6)]^(3+)` is found to be 240 kJ `mol^(-1)` then predict the colour of the complex using the following tab

Description : The value of `Delta_(0) " for " [Ti(H_(2)O)_(6)]^(3+)` is found to be 240 kJ `mol^(-1)` then predict the colour of the complex using the following tab

Last Answer : The value of `Delta_(0) " for " [Ti(H_(2)O)_(6)]^(3+)` is found to be 240 kJ `mol^(-1)` then ... -34)J-sec,N_(A)=6xx10^(23)c=3xx10^(8) m//sec)`

Explain the following , giving reason . (a) Transition metal and many of their compounds show paramagnetic behavior (b) The enthalpy of atomisation of the transition metal is high . (c) The Transition metal generally form coloured compounds .

Description : Explain the following , giving reason . (a) Transition metal and many of their compounds show paramagnetic behavior (b) The enthalpy of atomisation of the transition metal is high . (c) The Transition metal generally form coloured compounds . 

Last Answer : Ans(a) It is due to unpaired electrons . (b) It is due to strong metallic bond and additional Covalent bonds due to presence of unpaired electron in d- orbitals (c) It is due to presence of unpaired electrons ,they undergo d-d transition . 

Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important f

Description : Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important f

Last Answer : Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature ... H_(2)O lt H_(2)Te -` boiling point

Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important f

Description : Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important f

Last Answer : Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ... oxides are usually covalent and acidic in nature

Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important f

Description : Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important f

Last Answer : Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature ... H_(2)O lt H_(2)Te -` boiling point

Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important f

Description : Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important f

Last Answer : Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ... oxides are usually covalent and acidic in nature

The chemical reactivity of noble gases involves the loss of electrons and hence it can form compounds with highly electronegative elements like F and

Description : The chemical reactivity of noble gases involves the loss of electrons and hence it can form compounds with highly electronegative elements like F and

Last Answer : The chemical reactivity of noble gases involves the loss of electrons and hence it can form compounds with ... )d` and linear D. `sp` and linear

The chemical reactivity of noble gases involves the loss of electrons and hence it can form compounds with highly electronegative elements like F and

Description : The chemical reactivity of noble gases involves the loss of electrons and hence it can form compounds with highly electronegative elements like F and

Last Answer : The chemical reactivity of noble gases involves the loss of electrons and hence it can form compounds with highly ... D. `sp^(3)d^(2)` & 2

The chemical reactivity of noble gases involves the loss of electrons and hence it can form compounds with highly electronegative elements like F and

Description : The chemical reactivity of noble gases involves the loss of electrons and hence it can form compounds with highly electronegative elements like F and

Last Answer : The chemical reactivity of noble gases involves the loss of electrons and hence it can form compounds with highly ... A. zero B. 2 C. 6 D. 8

Among the following, the number of elements showing only one non-zero oxidation state is: O,Cl,F,N,P,Sn,Tl,Na,Ti

Description : Among the following, the number of elements showing only one non-zero oxidation state is: O,Cl,F,N,P,Sn,Tl,Na,Ti

Last Answer : Among the following, the number of elements showing only one non-zero oxidation state is: O,Cl,F,N,P,Sn,Tl,Na,Ti

If a three hinged parabolic arch, (span l, rise h) is carrying a uniformly distributed load w/unit length over the entire span, (A) Horizontal thrust is wl2 /8h (B) S.F. will be zero throughout (C) B.M. will be zero throughout (D) All the above

Description : If a three hinged parabolic arch, (span l, rise h) is carrying a uniformly distributed load w/unit  length over the entire span,  (A) Horizontal thrust is wl2 /8h (B) S.F. will be zero throughout  (C) B.M. will be zero throughout  (D) All the above 

Last Answer : (D) All the above 

The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

Description : The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

Last Answer : The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g) ... mol^(-1)` D. `-92.0 KJ mol^(-1)`

The enthalpy changes of formation of the gaseous oxides of nitrogen `(N_(2)O` and `NO)` are positive because of `:`

Description : The enthalpy changes of formation of the gaseous oxides of nitrogen `(N_(2)O` and `NO)` are positive because of `:`

Last Answer : The enthalpy changes of formation of the gaseous oxides of nitrogen `(N_(2)O` and `NO)` are positive ... The tendency of oxygen to form `O^(2-)`

The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 kJ / mol respectively. For the reaction `N_(2)

Description : The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 kJ / mol respectively. For the reaction `N_(2)

Last Answer : The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 ... is A. `-212` B. `+212` C. `+778` D. `-778`

Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` is

Description : Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` is

Last Answer : Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. ... kJ mol^(-1)` D. `-245 kJ mol^(-1)`

Standard enthalpy of formation of ethane is -84.7kjper mol calculate the enthalpy change for the formation of 0.1kg ethane

Description : Standard enthalpy of formation of ethane is -84.7kjper mol calculate the enthalpy change for the formation of 0.1kg ethane

Last Answer : Standard enthalpy of formation of ethane is -84.7kjper mol calculate the enthalpy change for the formation of 0.1kg ethane

Which reaction shows that the enthalpy of formation of H2S is Hf -20.6 kJmol?

Description : Which reaction shows that the enthalpy of formation of H2S is Hf -20.6 kJmol?

Last Answer : H2(g) + S(s) —> H2S + 20.6 kJ

What is the sign of enthalpy of formation of highly stable compound?

Description : What is the sign of enthalpy of formation of highly stable compound?

Last Answer : Ans. Negative.

Write an expression in the form of chemical equation for the standard enthalpy of formation (∆Hf) of CO (g).

Description : Write an expression in the form of chemical equation for the standard enthalpy of formation (∆Hf) of CO (g).

Last Answer : Ans. C(s) +1/2 O2(g) → CO (g)

Entropy change of the reaction, H2O (liquid) → H2O (gas), is termed as the enthalpy of (A) Solution (B) Vaporisation (C) Formation (D) Sublimation

Description : Entropy change of the reaction, H2O (liquid) → H2O (gas), is termed as the enthalpy of (A) Solution (B) Vaporisation (C) Formation (D) Sublimation

Last Answer : (B) Vaporisation

Assertion :- Absolute value of enthalpy can not be determined. Reason :- Enthalpy is defined as H = E + PV, and value of internal energy can not be de

Description : Assertion :- Absolute value of enthalpy can not be determined. Reason :- Enthalpy is defined as H = E + PV, and value of internal energy can not be de

Last Answer : Assertion :- Absolute value of enthalpy can not be determined. Reason :- Enthalpy is defined as H = ... . D. If both Assertion & Reason are false.

The enthalpy changes at 298 K in successive breaking of `O-H` bonds of water, are `H_(2)O(g) rarr H(g)+OH(g),DeltaH=498kJ mol^(-1)` `OH(g) rarr H(g)+O

Description : The enthalpy changes at 298 K in successive breaking of `O-H` bonds of water, are `H_(2)O(g) rarr H(g)+OH(g),DeltaH=498kJ mol^(-1)` `OH(g) rarr H(g)+O

Last Answer : The enthalpy changes at 298 K in successive breaking of `O-H` bonds of water, are `H_(2)O(g) rarr H(g)+OH(g), ... KJ mol"^(-1)` D. `463 "KJ mol"^(-1)`

The enthalpy change `(Delta H)` for the reaction `N_(2) (g)+3H_(2)(g) rarr 2NH_(3)(g)` is `-92.38 kJ` at `298 K`. What is `Delta U` at `298 K`?

Description : The enthalpy change `(Delta H)` for the reaction `N_(2) (g)+3H_(2)(g) rarr 2NH_(3)(g)` is `-92.38 kJ` at `298 K`. What is `Delta U` at `298 K`?

Last Answer : The enthalpy change `(Delta H)` for the reaction `N_(2) (g)+3H_(2)(g) rarr 2NH_(3)(g)` is `-92.38 kJ` at ` ... -87.42 kJ` C. `-97.34 kJ` D. `-89.9 kJ`

High pressure steam is expanded adiabatically and reversibly through a well insulated turbine, which produces some shaft work. If the enthalpy change and entropy change across the turbine are represented by ΔH and ΔS respectively for this process: (A) Δ H = 0 and ΔS = 0 (B) Δ H ≠ 0 and ΔS = 0 (C) Δ H ≠ 0 and ΔS ≠ 0 (D) Δ H = 0 and ΔS ≠ 0

Description : High pressure steam is expanded adiabatically and reversibly through a well insulated turbine, which produces some shaft work. If the enthalpy change and entropy change across the turbine are represented by ΔH and ΔS respectively for this process: ... (C) Δ H ≠ 0 and ΔS ≠ 0 (D) Δ H = 0 and ΔS ≠ 0

Last Answer : (B) Δ H ≠ 0 and ΔS = 0

Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Description : Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Last Answer : 3300 lbf-ft/lbm

Metalloids (A) Are good conductor of heat & electricity (B) Act as electron donors with metals & as electron acceptor with non metals (C) Are not necessarily solids at room temperature (D) Are compounds that exhibit both metallic & non-metallic properties to some extent and are exemplified by elements like germanium, silicon & boron

Description : Metalloids (A) Are good conductor of heat & electricity (B) Act as electron donors with metals & as electron acceptor with non metals (C) Are not necessarily solids at room temperature ... & non-metallic properties to some extent and are exemplified by elements like germanium, silicon & boron

Last Answer : D) Are compounds that exhibit both metallic & non-metallic properties to some extent and are exemplified by elements like germanium, silicon & boron

If water is formed from `H^(+)` ions and `OH^(-)` the heat of formation of water is :

Description : If water is formed from `H^(+)` ions and `OH^(-)` the heat of formation of water is :

Last Answer : If water is formed from `H^(+)` ions and `OH^(-)` the heat of formation of water is : A. `-13.7` ... . 13.7 Kcal C. `-63.4` Kcal D. More data required

What is the standard enthalpy change at 298 K for the following reaction? `CO_(2) (g)+ C("diamond") rarr 2CO(g)` Given : `DeltaH_(f)^(@)(CO,g) = - 110

Description : What is the standard enthalpy change at 298 K for the following reaction? `CO_(2) (g)+ C("diamond") rarr 2CO(g)` Given : `DeltaH_(f)^(@)(CO,g) = - 110

Last Answer : What is the standard enthalpy change at 298 K for the following reaction? `CO_(2) (g)+ C("diamond") rarr 2CO(g ... `+172.5 kJ//mol` D. `170.5 kJ//mol`

Ionization enthalpy of Li is 520 kJ mol^(-1) while that of F is 1681 kJ mol^(-1) . Explain.

Description : Ionization enthalpy of Li is 520 kJ mol^(-1) while that of F is 1681 kJ mol^(-1) . Explain.

Last Answer : Ionization enthalpy of Li is 520 kJ mol-1 while that of F is 1681 kJ mol-1. Explain.

From each set, choose the element with largest ionization enthalpy and explain your answer. (a) F, O, N (b) Mg, P, Ar (c) B, Al, Ga

Description : From each set, choose the element with largest ionization enthalpy and explain your answer. (a) F, O, N (b) Mg, P, Ar (c) B, Al, Ga

Last Answer : Ans. (a). The element fluorine F has the largest ionization enthalpy because in general, it increases along the period. These elements belong to the second period in the order is: N, O, F. Therefore, ... the order of B, Al, Ga. Therefore, the first element B is expected to have the largest value.

Arrange the following order of the property indicated: (a). F, Cl, Br and I (negative electron gain enthalpy) (b). Mg, Al, Si and Na (ionization enthalpy) (c). C, N, O and F (second ionization enthalpy)

Description : Arrange the following order of the property indicated: (a). F, Cl, Br and I (negative electron gain enthalpy) (b). Mg, Al, Si and Na (ionization enthalpy) (c). C, N, O and F (second ionization enthalpy)

Last Answer : Ans. (a). I

Which out of F or Cl has a more negative electron gain enthalpy?

Description : Which out of F or Cl has a more negative electron gain enthalpy?

Last Answer : Ans. Cl has more negative electron gain enthalpy. It is because there is more inter electronic repulsion between valence electrons of F, due to smaller size than Cl.