Chemical Bonding & Molecular Structure

B is correct $\mathrm{KF}\mathrm{KF}$ $\mathrm{C}$ is correct $\underset{+4}{\mathrm{SnCl}_{4}}>\underset{+2}{\mathrm{SnCl}_{2}} ; \mathrm{CuCl}>\mathrm{NaCl}$ $\mathrm{E}$ is correct $\mathrm{KF}\mathrm{NaCl}$

Let's analyze each compound separately: 1.

$\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}$: This compound is also known as pyrophosphoric acid.

Its structure has two phosphorus atoms (P) connected by one oxygen atom (O), forming a P-O-P bond.

There are no other P-O-P bonds in its structure.

Number of P-O-P bonds = 1 2.

$\left(\mathrm{HPO}_{3}\right)_{3}$: This compound is also known as cyclotriphosphoric acid.

It has three phosphorus atoms and three oxygen atoms, arranged in a cyclic structure with each P atom connected to two O atoms, forming a six-membered ring.

In this structure, there are three P-O-P bonds.

Number of P-O-P bonds = 3 3.

$\mathrm{P}_{4} \mathrm{O}_{10}$: This compound is also known as phosphorus pentoxide.

Its structure consists of four phosphorus atoms (P) and ten oxygen atoms (O) connected in a cyclic arrangement.

In the structure, each phosphorus atom is connected to four oxygen atoms, with two terminal oxygen atoms and two bridging oxygen atoms.

There are six P-O-P bonds in the P4O10 structure.

So, the number of P-O-P bonds in $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7},\left(\mathrm{HPO}_{3}\right)_{3}$, and $\mathrm{P}_{4} \mathrm{O}_{10}$ are 1, 3, and 6, respectively.

The correct answer is (B) and (C).

(A) NF3 molecule has a trigonal pyramidal structure, not a trigonal planar structure.

Therefore, statement (A) is false.

(B) The bond length of $\mathrm{N}_{2}$ is shorter than that of $\mathrm{O}_{2}$ due to the smaller atomic radius of nitrogen compared to oxygen.

Therefore, statement (B) is true.

(C) Isoelectronic molecules or ions have the same number of electrons and hence the same bond order.

Therefore, statement (C) is true.

(D) The dipole moment of water ($\mathrm{H}_{2}\mathrm{O}$) is higher than that of $\mathrm{H}_{2}\mathrm{S}$ due to the greater electronegativity of oxygen compared to sulfur.

Therefore, statement (D) is false.

Statement I: Both SO₂ and H₂O have a V-shaped (or bent) molecular structure.

SO₂ has a central sulfur atom with two oxygen atoms and one lone pair, while H₂O has a central oxygen atom with two hydrogen atoms and two lone pairs.

In both cases, the repulsion between the lone pairs and the bonded electron pairs leads to a V-shaped structure.

Statement II: The bond angle in H₂O is approximately 104.5° due to the tetrahedral arrangement of electron pairs (including the lone pairs) around the central oxygen atom.

In SO₂, the bond angle is approximately 119°.

The presence of a lone pair and the double bonds in SO₂ results in a different electron distribution, leading to a larger bond angle compared to H₂O.

Let's analyze each option: Option A: $\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{2-}$ The bond order of $\mathrm{O}_{2}$ is 2 and it is paramagnetic.

When two electrons are added to form $\mathrm{O}_{2}^{2-}$, the bond order becomes 1 (decreases) and it becomes diamagnetic.

This option does not meet the required conditions.

Option B: $\mathrm{N}_{2} \rightarrow \mathrm{N}_{2}^{+}$ The bond order of $\mathrm{N}_{2}$ is 3 and it is diamagnetic.

When one electron is removed to form $\mathrm{N}_{2}^{+}$, the bond order becomes 2.5 (decreases) and it remains diamagnetic.

This option does not meet the required conditions.

Option C: $\mathrm{NO} \rightarrow \mathrm{NO}^{+}$ The bond order of $\mathrm{NO}$ is 2.5 and it is paramagnetic.

When one electron is removed to form $\mathrm{NO}^{+}$, the bond order becomes 3 (increases) and it becomes diamagnetic.

This option meets the required conditions.

Option D: $\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{+}$ The bond order of $\mathrm{O}_{2}$ is 2 and it is paramagnetic.

When one electron is removed to form $\mathrm{O}_{2}^{+}$, the bond order becomes 2.5 (increases) and it remains paramagnetic.

This option does not meet the required conditions.

Therefore, the correct answer is $\mathrm{NO} \rightarrow \mathrm{NO}^{+}$.

$\mathrm{ClF}_5$ is actually a colorless liquid at room temperature with a square pyramidal geometry.

We want to find which of the given diatomic species (among $\mathrm{O}_2^+$, $\mathrm{O}_2^-$, $\mathrm{N}_2^+$, and $\mathrm{NO}^+$) has the same bond order and the same magnetic property (dia- or paramagnetic) as the acetylide ion, $\mathrm{C}_2^{2-}$.

1.

Electron Count and MO Considerations (a) Acetylide Ion, $\mathbf{C_2^{2-}}$ A neutral $\mathrm{C}_2$ molecule would have $2 \times 6 = 12$ electrons.

In $\mathrm{C}_2^{2-}$, there are 2 extra electrons, giving a total of $12 + 2 = 14$ electrons. (b) Matching Total Electrons Among the Options Let us see how many electrons are in each species: $\mathrm{O}_2$ has $2 \times 8 = 16$ electrons $\mathrm{O}_2^+$ has $16 - 1 = 15$ electrons $\mathrm{O}_2^-$ has $16 + 1 = 17$ electrons $\mathrm{N}_2$ has $2 \times 7 = 14$ electrons $\mathrm{N}_2^+$ has $14 - 1 = 13$ electrons $\mathrm{NO}$ has $7 + 8 = 15$ electrons $\mathrm{NO}^+$ has $15 - 1 = 14$ electrons Hence the only diatomic species in the list that has 14 electrons (like $\mathrm{C}_2^{2-}$) is $\mathrm{NO}^+$.

2.

Bond Order and Magnetic Property (a) $\mathrm{C}_2^{2-}$ (14 Electrons) From molecular orbital (MO) theory for 14-electron diatomics (similar to $\mathrm{N}_2$ with 14e), we generally get: Bond order = 3 All electrons paired in MOs → diamagnetic. (b) $\mathrm{NO}^+$ (also 14 Electrons) Similarly, $\mathrm{NO}^+$ also has 14 electrons and, by the same reasoning in MO theory, has: Bond order = 3 Diamagnetic (no unpaired electrons).

3.

Conclusion Among the given choices: $\mathrm{O}_2^+$ has 15 electrons (not 14) $\mathrm{O}_2^-$ has 17 electrons $\mathrm{N}_2^+$ has 13 electrons $\mathrm{NO}^+$ has 14 electrons Therefore, the species with the same bond order (3) and the same magnetic property (diamagnetic) as $\mathrm{C}_2^{2-}$ is: $\boxed{\mathrm{NO}^{+} \text{ (Option D).}}$

To determine the number of molecules/ions in which the central atom is involved in

hybridization, we must analyze the hybridization state for each central atom.

Hybridization is typically determined by the number of sigma bonds and lone pairs on the central atom.

Let's evaluate each molecule/ion: 1.

: The central atom is nitrogen (N). In

, nitrogen forms three sigma bonds with oxygen atoms and has no lone pairs.

Using the formula for hybridization: Number of hybrid orbitals = number of sigma bonds + number of lone pairs For

:

hybrid orbitals. Therefore, nitrogen is

hybridized. 2.

: The central atom is boron (B). In

, boron forms three sigma bonds with chlorine atoms and has no lone pairs. Using the same formula: For

:

hybrid orbitals. Therefore, boron is

hybridized. 3.

: The central atom is chlorine (Cl). In

, chlorine forms two sigma bonds with oxygen atoms and has two lone pairs. Therefore,: For

:

hybrid orbitals. Thus, chlorine is

hybridized. 4.

: The central atom is chlorine (Cl). In

, chlorine forms three sigma bonds with oxygen atoms and has one lone pair. Therefore,: For

:

hybrid orbitals. Thus, chlorine is

hybridized. Based on the analysis, the molecules/ions

and

have their central atoms involved in

hybridization. Therefore, the total number is 2. Thus, the correct option is Option A.

Let's analyze both statements: Statement (I): "A $\pi$ bonding MO has lower electron density above and below the inter-nuclear axis."

This statement is false.

In molecular orbital (MO) theory, a pi bond ($\pi$ bond) is formed by the sideways overlap of p-orbitals from two adjacent atoms.

The characteristic electron density of a $\pi$ bonding molecular orbital is concentrated above and below the inter-nuclear axis, not lower.

These regions of electron density are where the p-orbitals overlap and electrons are likely to be found.

This concentration of electron density above and below the inter-nuclear axis is what allows the $\pi$ bond to provide additional stability to the molecule, alongside any sigma ($\sigma$) bonds that may be present.

Statement (II): "The $\pi^*$ antibonding MO has a node between the nuclei."

This statement is true.

In a $\pi^*$ (pi-star) antibonding molecular orbital, there is indeed a nodal plane between the nuclei where the probability of finding electrons is essentially zero.

This node arises from the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive interference and a region of zero electron density in the bonding region.

The presence of this nodal plane is what makes the $\pi^*$ orbital antibonding—the electrons in this orbital actually serve to destabilize the bond between the two atoms.

Therefore, the correct answer is: Option D: Statement I is false but Statement II is true.

The correct answer is Option C: (A) is correct but (R) is not correct.

Assertion (A): $\mathrm{PH}_3$ has lower boiling point than $\mathrm{NH}_3$.

This assertion is true.

The boiling point of ammonia ($\mathrm{NH}_3$) is higher than that of phosphine ($\mathrm{PH}_3$).

Ammonia has a boiling point of about -33.34°C, whereas phosphine has a boiling point of about -87.7°C.

The reason for the higher boiling point of ammonia as compared to phosphine lies in the strength and type of intermolecular forces present in the substances.

Reason (R) : In liquid state $\mathrm{NH}_3$ molecules are associated through Vander Waals' forces, but $\mathrm{PH}_3$ molecules are associated through hydrogen bonding.

This reason is incorrect.

Ammonia ($\mathrm{NH}_3$) can form hydrogen bonds due to the presence of a highly electronegative nitrogen atom bonded to hydrogen atoms.

This allows $\mathrm{NH}_3$ molecules to strongly associate with each other through hydrogen bonding, which is a much stronger intermolecular force than Van der Waals forces.

This hydrogen bonding is responsible for the relatively high boiling point of ammonia.

On the other hand, phosphine ($\mathrm{PH}_3$) does not form hydrogen bonds because the electronegativity difference between phosphorus and hydrogen is not significant enough to enable the formation of hydrogen bonds.

Instead, phosphine molecules are associated mainly through weaker Van der Waals forces, leading to a lower boiling point when compared to ammonia.

In conclusion, the assertion that $\mathrm{PH}_3$ has a lower boiling point than $\mathrm{NH}_3$ is correct, due to the hydrogen bonding present in $\mathrm{NH}_3$ and absent in $\mathrm{PH}_3$.

However, the reason provided is incorrect because it incorrectly states that $\mathrm{PH}_3$ forms hydrogen bonds and that $\mathrm{NH}_3$ is associated through Van der Waals forces.

It is actually the other way around, so the correct answer is that (A) is true but (R) is false.