Which of the following statement is not correct from the view point of molecular orbital theory? (i) Be2 is no

NCERT Exemplar · Q19

Q. Which of the following statement is not correct from the view point of molecular orbital theory?

(i) Be2 is not a stable molecule.

(ii) He2 is not stable but He2^+ is expected to exist.

(iii) Bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period.

(iv) The order of energies of molecular orbitals in N2 molecule is σ2s < σ2s < σ2p_z < (π2p_x = π2p_y) < (π2p_x = π2p_y) < σ2p_z

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Concept understanding — Lewis Dot Structures

Why do we need Lewis dot structures?

Atoms are held together in molecules by chemical bonds — but what exactly is a bond? In the early 20th century, Gilbert N. Lewis realised that the key lies in the valence electrons (the outermost electrons). He noticed that atoms of noble gases (like Ne, Ar) are extremely stable and unreactive, and they all have 8 electrons in their outermost shell (except helium, which has 2). This led to the octet rule: atoms tend to gain, lose, or share electrons to achieve a full outer shell of 8 electrons (or 2 for hydrogen).

Lewis dot structures are simply a shorthand picture of this idea. They show:

Which atoms are connected to which
How many valence electrons each atom contributes
How those electrons are arranged as bonding pairs (shared) or lone pairs (unshared)

The precise statement

A Lewis dot structure (or electron dot structure) represents the valence electrons of an atom or molecule using dots placed around the element's symbol. Each dot stands for one valence electron. Shared pairs (bonds) are shown as lines, and unshared pairs as pairs of dots.

For a single atom, you write the element symbol and place dots on its four sides (top, bottom, left, right) — up to 8 dots. The order of filling doesn't matter for the final picture, but conventionally you place one dot on each side first, then pair them up.

For example:

Carbon (group 14, 4 valence electrons): $\cdot \text{C} \cdot$ (four single dots)
Oxygen (group 16, 6 valence electrons): $\cdot \ddot{\text{O}} \cdot$ (two single dots and two pairs)

How to draw a Lewis structure for a molecule

Here's the step-by-step method you'll use in exams:

Count total valence electrons — add up valence electrons from all atoms. For ions, add 1 electron for each negative charge, subtract 1 for each positive charge.
Identify the central atom — usually the least electronegative element (not hydrogen or fluorine). Place it in the centre.
Connect atoms with single bonds — each bond uses 2 electrons. Subtract these from your total.
Complete octets of outer atoms — place remaining electrons as lone pairs on terminal atoms (except hydrogen, which only needs 2).
Place leftover electrons on the central atom — if any remain.
If the central atom has fewer than 8 electrons, form multiple bonds — move lone pairs from outer atoms to create double or triple bonds until the central atom has an octet.

Watch out

A common mistake: forgetting that hydrogen only needs 2 electrons (a duet), not 8. Never put more than 2 electrons around H.

A concrete example: water (H₂O)

Total valence electrons: O has 6, each H has 1 → $6 + 1 + 1 = 8$ electrons.
Central atom: oxygen (least electronegative after H).
Connect: O—H bonds (2 bonds × 2 electrons = 4 electrons used).
Remaining: $8 - 4 = 4$ electrons → place as two lone pairs on oxygen.
Check: O has 2 bonds (4 electrons) + 2 lone pairs (4 electrons) = 8. Each H has 1 bond (2 electrons) = 2. Done.

The structure: $\text{H} - \ddot{\text{O}} - \text{H}$

What the structure tells you

Once drawn, a Lewis structure reveals:

Bond order (single, double, triple)
Lone pairs (which affect molecular shape and reactivity)
Formal charge (a bookkeeping tool to check which structure is most stable) …

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The energy ordering in statement (D) is the one valid for $\text{O}_2$ / $\text{F}_2$ , not for $\text{N}_2$ , so (D) is the incorrect statement — option (D).

Checking each statement against molecular orbital theory:

(A) $\text{Be}_2$ is not stable — correct. Its configuration gives bonding and antibonding $2s$ electrons that cancel, so bond order $=\tfrac{1}{2}(4-4)=0$ ; the molecule does not exist.
(B) $\text{He}_2$ unstable but $\text{He}_2^{+}$ expected — correct. $\text{He}_2$ : bond order $=\tfrac{1}{2}(2-2)=0$ ; $\text{He}_2^{+}$ : bond order $=\tfrac{1}{2}(2-1)=0.5$ , so it can exist.
(C) $\text{N}_2$ has the maximum bond strength among period-2 homonuclear diatomics — correct. $\text{N}_2$ has bond order $3$ , the highest of the series. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{BeCl}_2$
›Reveal solutionSolution
The key is to identify which pair contains one molecule with an unpaired electron (odd-electron species) and another where the central atom exceeds the octet. The correct pair is NO (odd electron) and PF₅ (expanded octet), which corresponds to option (B).

Concept & Intuition

We need to check two distinct chemical concepts in each pair:

Odd-electron molecule: A molecule with an unpaired electron (a radical), often resulting from an odd total number of valence electrons. Common examples: NO, NO₂, ClO₂.

Expanded octet: A molecule where the central atom has more than 8 electrons in its valence shell, possible only for elements in period 3 or beyond (e.g., P, S, Cl) because they have available d-orbitals.

We’ll examine each pair systematically.

Pair (A): BeCl₂ & HNO₃

BeCl₂: Beryllium has only 2 valence electrons; it forms two bonds, giving it 4 electrons around it — a reduced octet, not odd-electron.

HNO₃: Nitrogen has 5 valence electrons, forms three bonds and has no lone pairs (in its typical Lewis structure), giving it a full octet. No unpaired electrons.

Neither fits. ✗

Pair (B): NO & PF₅

NO (nitric oxide): Total valence electrons = 5 (N) + 6 (O) = 11. An odd number means at least one unpaired electron. The Lewis structure shows N with 7 electrons and O with 8, leaving one unpaired electron on N. ✓ Odd-electron molecule.

PF₅ (phosphorus pentafluoride): Phosphorus (period 3) has 5 valence electrons; it forms 5 single bonds, giving it 10 electrons around it — an expanded octet. ✓

This pair satisfies both conditions. ✓

Pair (C): BCl₃ & NO₂

BCl₃: Boron has 3 valence electrons, forms three bonds, giving only 6 electrons — an incomplete octet, not odd-electron.

NO₂: Total valence electrons = 5 + 12 = 17 (odd), so it is an odd-electron molecule. But BCl₃ is not odd-electron. ✗ …
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KCET 2022Set B-31 markMCQ
$O_2: 1\sigma, 1\pi; C_2: 0\sigma, 2\pi$
›Reveal solutionSolution
Fill the MO diagrams: $O_2$ ends up $1\sigma + 1\pi$ , while $C_2$ — because $\pi_{2p}$ lies below $\sigma_{2p_z}$ for Li–N — ends up with $2\pi$ and no net $\sigma$ bond.

1. The crucial MO ordering rule

For the second-period diatomics there are two orbital orderings:

$B_2$ to $N_2$ (small $2s$ – $2p$ energy gap $\Rightarrow$ strong $s$ – $p$ mixing):

$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \boxed{\pi_{2p_x} = \pi_{2p_y}} < \sigma_{2p_z} < \pi^*_{2p} < \sigma^*_{2p}$

$O_2$ , $F_2$ , $Ne_2$ (large gap, no mixing):

$\dots < \sigma^*_{2s} < \boxed{\sigma_{2p_z}} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p} < \sigma^*_{2p}$
The $\pi/\sigma$ swap between $N_2$ and $O_2$ is the whole key to this question.

2. Oxygen molecule, $O_2$ (16 electrons)
$\sigma_{1s}^2\,\sigma^*_{1s}{}^2\,\sigma_{2s}^2\,\sigma^*_{2s}{}^2\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\pi^*_{2p_x}{}^1\,\pi^*_{2p_y}{}^1$ $\text{Bond order} = \tfrac{1}{2}(N_b - N_a) = \tfrac{1}{2}(10 - 6) = 2$
Counting the net bonds:

$\sigma_{2p_z}^2$ is unopposed $\Rightarrow$ one full $\sigma$ bond.

The four $\pi$ bonding electrons are partly cancelled by the two electrons in $\pi^*$ (one in each, by Hund's rule — this is also why $O_2$ is paramagnetic). Net $\pi$ contribution $=\tfrac12(4-2) = 1$ $\Rightarrow$ one $\pi$ bond (really two "three-electron" half- $\pi$ bonds).

$\therefore\ O_2:\ 1\sigma,\ 1\pi \quad(\text{total bond order }2)$
3. Carbon molecule, $C_2$ (12 electrons)

Using the $B_2$ – $N_2$ ordering, where $\pi_{2p}$ comes first:
$\sigma_{1s}^2\,\sigma^*_{1s}{}^2\,\sigma_{2s}^2\,\sigma^*_{2s}{}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2$
$\text{Bond order} = \tfrac{1}{2}(8 - 4) = 2$ …
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KCET 2020Set A-11 markMCQ
$+1$
›Reveal solutionSolution
Draw the resonance structure of $\text{O}_3$ and apply $F.C. = V - L - \tfrac{1}{2}B$ to the central atom.

Step 1 — The formal-charge formula.

For an atom in a Lewis structure,
$F.C. = V - L - \frac{1}{2}B$
where $V$ = valence electrons of the free atom, $L$ = number of non-bonding (lone-pair) electrons, $B$ = number of bonding electrons around that atom.

Step 2 — Lewis structure of ozone.

Ozone is bent; one resonance structure is
$\text{O}_{(1)} = \text{O}_{(2)} - \text{O}_{(3)}^{\;-}$
The central oxygen $\text{O}_{(2)}$ therefore:

forms a double bond to one terminal O (4 bonding electrons)

forms a single bond to the other terminal O (2 bonding electrons)

carries one lone pair (2 non-bonding electrons)

So $V=6$ , $L=2$ , $B=4+2=6$ .

Step 3 — Compute.
$F.C.(\text{central O}) = 6 - 2 - \frac{1}{2}(6) = 6 - 2 - 3 = +1$
Step 4 — Sanity check with the whole molecule. …
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KCET 2019Set A-11 markMCQ

I_3^-, H_2O

›Reveal solution

Draw each species by VSEPR: count the central atom's valence electrons, subtract those used in bonding, and convert the remainder into lone pairs; then find the pair in which both central atoms are lone-pair rich.

Step 1 — The counting rule.

For a central atom $A$ , the number of electron domains is

\text{domains} = \sigma\text{-bonds} + \text{lone pairs}, \qquad \text{lone pairs} = \frac{V + C - B}{2}

where $V$ = valence electrons of $A$ , $C$ = charge contribution, $B$ = number of $\sigma$ -bonded atoms (each single bond uses one of $A$ 's electrons).

Step 2 — Species by species.

Species	Central atom	Valence e⁻ used	Lone pairs	Shape
$I_3^-$	I	$7+1 = 8$ ; 2 bonds	$(8-2)/2 = \mathbf{3}$	linear (trigonal bipyramidal e⁻ geometry)
$H_2O$	O	6; 2 bonds	$(6-2)/2 = \mathbf{2}$	bent
$XeF_4$	Xe	8; 4 bonds	$(8-4)/2 = \mathbf{2}$	square planar
$NH_3$	N	5; 3 bonds	$(5-3)/2 = \mathbf{1}$	pyramidal
$NF_3$	N	5; 3 bonds	$(5-3)/2 = \mathbf{1}$	pyramidal
$SO_4^{2-}$	S	all used in 4 bonds	$\mathbf{0}$	tetrahedral
$H_2S$	S	6; 2 bonds	$(6-2)/2 = \mathbf{2}$	bent

Step 3 — Test each printed pair.

(A) $I_3^-$ (3 lp) and $H_2O$ (2 lp) — both central atoms are lone-pair bearing, with more than one lone pair each. …

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Q.Which of the following statement is not correct from the view point of molecular orbital theory?

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