Question

question 2
for each of the following molecules identify the hybridization.
n₂h₂
a. ○ sp
b. ○ sp²
c. ○ sp³
d. ○ sp³d
e. ○ sp³d²
question 3
clf₅
a. ○ sp
b. ○ sp²
c. ○ sp³
d. ○ sp³d
e. ○ sp³d²

Explanation:

Question 2

The formula for calculating the hybridization is \( H=\frac{1}{2}(V + M - C + A) \), where \( V \) is the number of valence electrons of the central atom, \( M \) is the number of monovalent atoms, \( C \) is the charge of the cation, and \( A \) is the charge of the anion. For \( N_2H_2 \), each \( N \) atom is the central atom. The valence electrons of \( N \) (\( V = 5 \)), monovalent atoms (\( M = 2 \) for each \( N \) considering \( H \) atoms), \( C = 0 \), \( A = 0 \). So \( H=\frac{1}{2}(5 + 2)=3.5\approx 3 \) (we take the integer part). But another way is to look at the structure: \( N_2H_2 \) has a double - bond between \( N \) atoms (\( H - N=N - H \)). The \( N \) atom has 3 electron - pair domains (2 single bonds with \( H \) and 1 double bond with \( N \)). According to the hybridization rules, 3 electron - pair domains correspond to \( sp^{2} \) hybridization.

For \( ClF_5 \), the central atom is \( Cl \). The valence electrons of \( Cl \) (\( V=7 \)), the number of monovalent \( F \) atoms (\( M = 5 \)), \( C = 0 \), \( A = 0 \). Using the formula \( H=\frac{1}{2}(V + M)=\frac{1}{2}(7 + 5)=6 \). According to the hybridization rules, 6 electron - pair domains correspond to \( sp^{3}d^{2} \) hybridization.

Answer:

B. \( sp^{2} \)

Question 3