The hybridisation of Xe in `XeO_(3)` is

The hybridisation of Xe in `XeO_(3)` is
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The hybridisation of Xe in `XeO_(3)` is A. `sp^(2)` B. `sp^(3)d` C. `sp^(3)` D. `sp^(3)d^(2)`
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Among the following molecules, `(i)XeO_(3)(ii)XeOF_(4)(iii)XeF_(6)` those having same number of lone pairs on `Xe` are:

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In `XeO_(3)` and `XeF_(6)` the oxidation state of `Xe` is

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Oxidation state of Xe in `Ba_(2)[XeO_(6)]` is

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The increasing d-character in hybridisation of Xe in `XeF_(2), XeF_(4), XeF_(6)` is

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The hybridisation of Xe is `sp^(3)d^(2)` in

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`XeO_(3)` forms xenate ion in alkaline medium. `XeO_(3) +NaOH rarr Na[HXeO_(4)]` But the xenate ions slowly disproportionate in alkaline solution as `

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The structure of `XeO_(2)F_(2)` is

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The hydrolysis of `XeF_(6)` takes place in the following steps : `XeF_(6) rarr A rarr B rarr XeO_(3)`. Then the correct statement regarding A and B is

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Description : `XeO_(4)` contains :

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The shape of `XeO_(3)` molecule is

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`XeO_(3)` has

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The number of lone pairs of electrons present on Xe in `XeF_(2)` ?

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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.

Find the divergence of the vector F= xe -x i + y j – xz k a) (1 – x)(1 + e -x ) b) (x – 1)(1 + e -x ) c) (1 – x)(1 – e) d) (x – 1)(1 – e)

Description : Find the divergence of the vector F= xe -x i + y j – xz k a) (1 – x)(1 + e -x ) b) (x – 1)(1 + e -x ) c) (1 – x)(1 – e) d) (x – 1)(1 – e)

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Botanical Words Alphabetical List - XE

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What is hybridisation on the central atom of `SiO_(2)`.

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Which of the following pair have same hybridisation :-

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Which technique can be helpful in overcoming hybridisation barrier ?

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Somatic hybridisation is achieved through ?

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