What is the hybridization of the nitrogen ato | vedclass

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(B) To determine the hybridization,we calculate the steric number $(X)$ using the formula: $X = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of

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$sp, sp^2, sp^3$

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(B) To determine the hybridization,we calculate the steric number $(X)$ using the formula: $X = \frac{1}{2}(V + M - C + A)$,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.$1$. For $NO_2^+$: $V = 5$,$M = 0$,$C = 1$,$A = 0$. $X = \frac{1}{2}(5 + 0 - 1 + 0) = 2$. Hybridization is $sp$.$2$. For $NO_3^-$: $V = 5$,$M = 0$,$C = 0$,$A = 1$. $X = \frac{1}{2}(5 + 0 - 0 + 1) = 3$. Hybridization is $sp^2$.$3$. For $NH_4^+$: $V = 5$,$M = 4$,$C = 1$,$A = 0$. $X = \frac{1}{2}(5 + 4 - 1 + 0) = 4$. Hybridization is $sp^3$.Thus,the hybridization states are $sp, sp^2, sp^3$ respectively.

What is the hybridization of the nitrogen atomic orbitals in $NO_2^+$,$NO_3^-$,and $NH_4^+$ respectively?

Similar Questions

Which of the following is an incorrect relation based on Bent's rule?

Which of the following molecular geometries is not produced by $sp^3$ hybridisation?

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

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