Question

In which one of the following the central atom is sp\textsuperscript{3 hybridized?}

• A. NH\textsubscript{4}\textsuperscript{+}
• B. BF\textsubscript{3}
• C. SF\textsubscript{6}
• D. PCl\textsubscript{5}
• E. XeF\textsubscript{4}

Hint:

Carbon, Nitrogen, and Oxygen in saturated, single-bonded environments (like $CH_4, NH_4^+, H_2O$) are almost always sp\textsuperscript{3} hybridized.

Answer 1

The Correct Option is A

Concept: The hybridization of a central atom is determined by its steric number (sum of bond pairs and lone pairs).
• sp\textsuperscript{3 Hybridization}: Occurs when the steric number is 4.
• sp\textsuperscript{2 Hybridization}: Steric number 3.
• sp\textsuperscript{3d Hybridization}: Steric number 5.
• sp\textsuperscript{3d\textsuperscript{2} Hybridization}: Steric number 6.

Step 1: Calculate steric number for each option.
• NH\textsubscript{4\textsuperscript{+}}: $N$ has 5 valence electrons. $+4H$ atoms, $-1(\text{charge}) = 8 / 2 = \mathbf{4}$. (4 BP + 0 LP).
• BF\textsubscript{3: $B$ has 3 valence electrons. $+3F$ atoms $= 6 / 2 = \mathbf{3}$.
• SF\textsubscript{6: $S$ has 6 valence electrons. $+6F$ atoms $= 12 / 2 = \mathbf{6}$.
• PCl\textsubscript{5: $P$ has 5 valence electrons. $+5Cl$ atoms $= 10 / 2 = \mathbf{5}$.
• XeF\textsubscript{4: $Xe$ has 8 valence electrons. $+4F$ atoms $= 12 / 2 = \mathbf{6}$.

Step 2: Match with sp\textsuperscript{3}. Only the Ammonium ion ($NH_4^+$) has a steric number of 4, which corresponds to $sp^3$ hybridization. It has a tetrahedral geometry.