Identify the ideal bond angle $109^\circ 28'$

The angle $109^\circ 28'$ is the ideal tetrahedral angle: $\theta = \cos^{-1}\left(-\frac{1}{3}\right) \approx 109.5^\circ$ So we need species having tetrahedral geometry with no lone-pair distortion.

Check each species

(i) $NH_3$

• Central atom: N
• Valence shell: 3 bond pairs + 1 lone pair
• Electron-pair geometry: tetrahedral
• Molecular shape: trigonal pyramidal
• Bond angle is about $107^\circ$, not $109^\circ 28'$

So, $NH_3$ is not correct.

(ii) $BF_4^-$

• Central atom: B
• Four $B-F$ bonds, no lone pair on B
• Shape: tetrahedral
• Bond angle: $109^\circ 28'$

So, $BF_4^-$ is correct.

(iii) $NH_4^+$

• Central atom: N
• Four $N-H$ bonds, no lone pair on N
• Shape: tetrahedral
• Bond angle: $109^\circ 28'$

So, $NH_4^+$ is correct.

(iv) $BF_3$

• Central atom: B
• Three bond pairs, no lone pair
• Shape: trigonal planar
• Bond angle: $120^\circ$

So, $BF_3$ is not correct.

(v) $NH_2^-$

• Central atom: N
• Two bond pairs + two lone pairs
• Electron-pair geometry: tetrahedral
• Molecular shape: bent
• Bond angle is much less than $109.5^\circ$

So, $NH_2^-$ is not correct.

Evaluate options

• A: $NH_3$, $BF_4^-$ → $NH_3$ incorrect
• B: $NH_4^+$, $BF_4^-$ → both correct
• C: $NH_3$, $BF_3$ → both incorrect
• D: $NH_2^-$, $BF_3$ → both incorrect

Final answer

The species having bond angle $109^\circ 28'$ are: $NH_4^+,\ BF_4^-$ Hence, the correct option is B.

Comparison with stored answer

Stored correct answer: B

My derived answer: B

Therefore, they agree.