Which of the two ions from the list given bel | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$NO_2^-$ and $NO_3^-$

Vedclass · Answer

(A) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,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.$1. NO_2^-: SN = \frac{1}{2} (5 + 0 - 0 + 1) = 3 \rightarrow sp^2$ hybridization.$2. NO_3^-: SN = \frac{1}{2} (5 + 0 - 0 + 1) = 3 \rightarrow sp^2$ hybridization.$3. NH_2^-: SN = \frac{1}{2} (5 + 2 - 0 + 1) = 4 \rightarrow sp^3$ hybridization.$4. NH_4^+: SN = \frac{1}{2} (5 + 4 - 1 + 0) = 4 \rightarrow sp^3$ hybridization.$5. SCN^-: SN = \frac{1}{2} (4 + 0 - 0 + 1) = 2 \rightarrow sp$ hybridization.Comparing the results,$NO_2^-$ and $NO_3^-$ both exhibit $sp^2$ hybridization.

Which of the two ions from the list given below have the geometry that is explained by the same hybridization of orbitals: $NO_2^-$,$NO_3^-$,$NH_2^-$,$NH_4^+$,$SCN^-$?

Similar Questions

In $CO_2$,$C$ is $sp$ hybridised using $s$ and $p_x$ orbitals. Then:

The type of hybridisation of boron in diborane $(B_2H_6)$ is:

The hybridisation of $C$ in diamond,graphite and ethyne is in the order

The percentage of $p$-character of the hybrid orbitals in graphite and diamond are respectively

What is the hybridization state of the central atom in the conjugate base of $NH_4^{+}$ ion?

Vedclass Test Series

Exam Paper Generator

Online Exam Module