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(A) To determine the hybridization,we use the formula: $	ext{Steric Number} = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electro

Vedclass · Accepted Answer

$A-R, B-S, C-P, D-R$

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(A) To determine the hybridization,we use the formula: $	ext{Steric Number} = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$(A)$ $ICl_4^-$: $	ext{Steric Number} = \frac{1}{2} (7 + 4 + 1) = 6$. Hybridization is $sp^3d^2$ $(R)$.$(B)$ $NO_3^-$: $	ext{Steric Number} = \frac{1}{2} (5 + 0 + 1) = 3$. Hybridization is $sp^2$ $(S)$.$(C)$ $PCl_4^+$: $	ext{Steric Number} = \frac{1}{2} (5 + 4 - 1) = 4$. Hybridization is $sp^3$ $(P)$.$(D)$ $SiF_6^{2-}$: $	ext{Steric Number} = \frac{1}{2} (4 + 6 + 2) = 6$. Hybridization is $sp^3d^2$ $(R)$.Thus,the correct matching is $A-R, B-S, C-P, D-R$.

Match the following:
| List-$I$ (Molecule/ion) | List-$II$ (Hybridisation) |
| :--- | :--- |
| $(A)$ $ICl_4^-$ | $(P)$ $sp^3$ |
| $(B)$ $NO_3^-$ | $(Q)$ $sp^3d$ |
| $(C)$ $PCl_4^+$ | $(R)$ $sp^3d^2$ |
| $(D)$ $SiF_6^{2-}$ | $(S)$ $sp^2$ |
| | $(T)$ $sp$ |

Similar Questions

Identify the compound which is polar and has the central atom in $sp^2$ hybridization.

In which of the following pairs do both the species have $sp^{3}$ hybridization?

In methane,the bond angle is $......^o$.

The compounds with $sp^2$ hybridized central atom among the following are:
$(A)$ $H_2CO_3$
$(B)$ $SiF_4$
$(C)$ $BF_3$
$(D)$ $HClO_2$

The number of species from the following in which the central atom uses $sp^3$ hybrid orbitals in its bonding is . . . . . . .
$NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3$

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Match the following:| List-$I$ (Molecule/ion) | List-$II$ (Hybridisation) || :--- | :--- || $(A)$ $ICl_4^-$ | $(P)$ $sp^3$ || $(B)$ $NO_3^-$ | $(Q)$ $sp^3d$ || $(C)$ $PCl_4^+$ | $(R)$ $sp^3d^2$ || $(D)$ $SiF_6^{2-}$ | $(S)$ $sp^2$ || | $(T)$ $sp$ |