Match List I and List II.
List I (Species)
(A) SF4
(B) IF5
(C) NO2+
(D) NH4+
List II (Hybrid orbital

Q: Why do we only count sigma bonds and not pi bonds for hybridisation?

Pi bonds are formed by unhybridised p-orbitals overlapping laterally. Only sigma bonds and lone pairs determine the basic electron geometry and require hybridised orbitals.

Q: How do I quickly find the number of valence electrons?

The number of valence electrons corresponds to the group number of the element in the periodic table (e.g., Group 16 elements like Oxygen and Sulfur have 6 valence electrons).

Q: Why does $NO_2^+$ have a linear shape?

Nitrogen in $NO_2^+$ has 4 valence electrons. It forms two double bonds with oxygen, using all 4 electrons. With 2 sigma bonds and 0 lone pairs, the steric number is 2, which corresponds to $sp$ hybridisation and a linear geometry.

Q: What is the difference between electron geometry and molecular shape?

Electron geometry considers both bond pairs and lone pairs (e.g., $SF_4$ has a trigonal bipyramidal electron geometry). Molecular shape only looks at the positions of the atoms (e.g., $SF_4$ has a see-saw shape due to the invisible lone pair).

Q: Why does the positive charge in $NH_4^+$ go to Nitrogen?

Nitrogen is less electronegative than the formal charge suggests in this context, but more importantly, it provides the empty orbital (after losing an electron) to accept the lone pair from ammonia when bonding with $H^+$.

Question

Match List I and List II.
\begin{array}{ll}
	ext{List I (Species)} & 	ext{List II (Hybrid orbitals)} \
	ext{(A) } SF_4 & 	ext{(i) } sp^3d^2 \
	ext{(B) } IF_5 & 	ext{(ii) } d^2sp^3 \
	ext{(C) } NO_2^+ & 	ext{(iii) } sp^3d \
	ext{(D) } NH_4^+ & 	ext{(iv) } sp^3 \
& 	ext{(v) } sp
\end{array}
Choose the correct answer from the options given below.

Options:

(A) (A)-(i), (B)-(ii), (C)-(v), (D)-(iii)

(B) (A)-(ii), (B)-(i), (C)-(iv), (D)-(v)

(C) (A)-(iii), (B)-(i), (C)-(v), (D)-(iv)

(D) (A)-(iv), (B)-(iii), (C)-(ii), (D)-(v)

Sigma Percentile · Accepted Answer

### The Magic of Steric Numbers

When it comes to predicting the shape and hybridisation of molecules, the **Steric Number** is your best friend. It is a simple yet incredibly powerful concept. The steric number ($H$) is calculated by adding the number of $\sigma$ (sigma) bonds formed by the central atom to the number of lone pairs residing on it.

$$H = 	ext{Number of } \sigma 	ext{ bonds} + 	ext{Number of lone pairs}$$

Let's apply this master formula to the four species given in our problem and decode their hybridisation states one by one.

### Analyzing Sulfur Tetrafluoride ($SF_4$)

Our first candidate is $SF_4$. The central atom is Sulfur, which belongs to Group 16 of the periodic table. This means it has **6 valence electrons**.

In $SF_4$, Sulfur forms 4 single bonds with 4 Fluorine atoms. Each single bond is a $\sigma$ bond, so we have 4 $\sigma$ bonds. These 4 bonds consume 4 of Sulfur's valence electrons.

We are left with $6 - 4 = 2$ electrons. These 2 electrons pair up to form exactly **1 lone pair**.

Now, let's calculate the steric number:
$$H = 4 	ext{ (bond pairs)} + 1 	ext{ (lone pair)} = 5$$

A steric number of 5 corresponds to **$sp^3d$ hybridisation**. While the base electron geometry is trigonal bipyramidal, the presence of one lone pair pushes the bonded atoms into a **see-saw** shape. Thus, (A) matches with (iii).

### Decoding Iodine Pentafluoride ($IF_5$)

Next up is $IF_5$. The central atom is Iodine, a halogen from Group 17, meaning it boasts **7 valence electrons**.

Iodine forms 5 single bonds with 5 Fluorine atoms, giving us 5 $\sigma$ bonds. This uses up 5 valence electrons.

Subtracting these from the total, we have $7 - 5 = 2$ electrons remaining, which constitutes **1 lone pair**.

Calculating the steric number:
$$H = 5 	ext{ (bond pairs)} + 1 	ext{ (lone pair)} = 6$$

A steric number of 6 dictates an **$sp^3d^2$ hybridisation**. The base geometry is octahedral, but with one lone pair, the molecular shape becomes **square pyramidal**. Therefore, (B) matches with (i).

### The Linear Nitronium Ion ($NO_2^+$)

Now, let's look at a cation: $NO_2^+$. The central atom is Nitrogen. Normally, Nitrogen (Group 15) has 5 valence electrons. However, the **positive charge** indicates the loss of one electron. So, Nitrogen here has $5 - 1 = 4$ valence electrons.

Nitrogen forms two double bonds with the two Oxygen atoms. A crucial rule in hybridisation is that **a double bond counts as only one $\sigma$ bond** (the other is a $\pi$ bond). So, we have 2 $\sigma$ bonds.

These two double bonds require all 4 of Nitrogen's valence electrons (2 for each Oxygen). This leaves **0 lone pairs**.

Calculating the steric number:
$$H = 2 	ext{ (bond pairs)} + 0 	ext{ (lone pairs)} = 2$$

A steric number of 2 corresponds to **$sp$ hybridisation**. This results in a perfectly **linear** geometry. Hence, (C) matches with (v).

### The Tetrahedral Ammonium Ion ($NH_4^+$)

Finally, we examine the ammonium ion, $NH_4^+$. Again, the central Nitrogen atom starts with 5 valence electrons, but the positive charge reduces this to **4 valence electrons**.

Nitrogen forms 4 single bonds with 4 Hydrogen atoms, giving us 4 $\sigma$ bonds. This consumes all 4 valence electrons, leaving **0 lone pairs**.

Calculating the steric number:
$$H = 4 	ext{ (bond pairs)} + 0 	ext{ (lone pairs)} = 4$$

A steric number of 4 means the hybridisation is **$sp^3$**. With no lone pairs to distort the shape, the molecule adopts a perfect **tetrahedral** geometry. Thus, (D) matches with (iv).

### Final Conclusion

By systematically applying the steric number formula, we have successfully determined the hybridisation for all four species:
- $SF_4 ightarrow sp^3d$
- $IF_5 ightarrow sp^3d^2$
- $NO_2^+ ightarrow sp$
- $NH_4^+ ightarrow sp^3$

This perfectly aligns with the sequence (A)-(iii), (B)-(i), (C)-(v), (D)-(iv), making option (c) the correct answer.

Select Answer:

Visualized Solution

$Matrix Match Analysis$

$The Steric Number$

$Analyzing S F_{4}$

$Hybridisation of S F_{4}$

$Analyzing I F_{5}$

$Hybridisation of I F_{5}$

$Analyzing N O_{2}^{+}$

$Hybridisation of N O_{2}^{+}$

$Analyzing N H_{4}^{+}$

$Hybridisation of N H_{4}^{+}$

$The Way Forward$

The Sigma Insight: Hybridisation and VSEPR Theory

The Magic of Steric Numbers

Analyzing Sulfur Tetrafluoride ( $S F_{4}$ )

Decoding Iodine Pentafluoride ( $I F_{5}$ )

The Linear Nitronium Ion ( $N O_{2}^{+}$ )

The Tetrahedral Ammonium Ion ( $N H_{4}^{+}$ )

Final Conclusion

Similar Questions

The hybridisations of the atomic orbitals of nitrogen in $NO_{2}^{-}$ , $NO_{2}^{+}$ and $NH_{4}^{+}$ respectively are

The species in which the N-atom is in a state of $s p$ hybridisation is

Among $[I_{3}]^{+}$ , $[SiO_{4}]^{4 -}$ , $SO_{2} Cl_{2}$ , $XeF_{2}$ , $SF_{4}$ , $ClF_{3}$ , $Ni(CO)_{4}$ , $XeO_{2} F_{2}$ , $[PtCl_{4}]^{2 -}$ , $XeF_{4}$ , and $SOCl_{2}$ , the total number of species having $s p^{3}$ hybridised central atom is ______.

The ion that has $s p^{3} d^{2}$ hybridisation for the central atom, is

List-I

List-II

Underlined carbon is $sp^{3}$ hybridised in

The pair of species having identical shapes for molecules of both species, is

In $CH_{2} = C = CH - CH_{3}$ molecule, the hybridisation of carbon 1, 2, 3 and 4 respectively are

The number of species below that have two lone pairs of electrons in their central atom is ……… . (Round off to the nearest integer) $SF_{4}, BF_{4}^{-}, ClF_{3}, AsF_{3}, PCl_{5}, BrF_{5}, XeF_{4}, SF_{6}$

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Matrix Match Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Steric Number

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing SF4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hybridisation of SF4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing IF5​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hybridisation of IF5​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing NO2+​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hybridisation of NO2+​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing NH4+​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hybridisation of NH4+​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward