P Block — NEET UG practice

1. Recall that the valence orbitals of an atom are all the orbitals in its outermost shell that are available for bonding; use this to check option A. Given aluminium (Al) has one $3s$, three $3p$, and five $3d$ orbitals available in its valence shell,

$$\text{Total valence orbitals of Al} = 1(3s) + 3(3p) + 5(3d) = 9$$

So option A, which states five valence orbitals, is incorrect.

2. Apply the same idea to beryllium, whose valence shell is the second shell containing only $s$ and $p$ orbitals.

$$\text{Total valence orbitals of Be} = 1(2s) + 3(2p) = 4$$

Hence, option C is incorrect since beryllium actually has four valence orbitals, not three.

3. Recall that maximum covalency depends on the number of orbitals an atom can use for bonding, including any accessible vacant $d$-orbitals; check magnesium.

$$\text{Maximum covalency of Mg} = 6$$

So option D is incorrect.

4. For boron, since its valence shell (second shell) has no $d$-orbitals, its maximum covalency is limited by the available $2s$ and $2p$ orbitals, allowing it to form up to four bonds (as seen in species like $[\text{BF}_4]^-$).

$$\text{Maximum covalency of B} = 4$$

Hence, the answer is B.

1. Recall that $p\pi-p\pi$ bonding occurs between atoms of similar size that use their $p$-orbitals to form multiple bonds; nitrogen is well known to do this with itself.

$$\text{N} \equiv \text{N} \ (\text{as in N}_2)$$

So option A is a correct statement.

2. Recall that elements with vacant $d$-orbitals in their valence shell can accept electron density back-donated from transition metals, forming a $d\pi-d\pi$ bond; phosphorus and arsenic both have accessible $d$-orbitals. Given this, $\text{P(C}_2\text{H}_5)_3$ and $\text{As(C}_6\text{H}_5)_3$ can indeed form $d\pi-d\pi$ bonds with transition metals, so option B is correct.

3. Recall catenation is the ability of atoms of the same element to link together via bonds; among Group 15 elements, phosphorus, arsenic, and antimony all show this property. So option C is a correct statement.

4. Check option D: forming a $d\pi-p\pi$ bond requires at least one atom to contribute empty $d$-orbitals. Since both nitrogen and oxygen belong to the second period,

$$\text{Valence orbitals of N and O: } 2s,\ 2p \quad (\text{no } 3d \text{ available})$$

neither atom has $d$-orbitals to offer, so nitrogen cannot form a $d\pi-p\pi$ bond with oxygen.

Hence, option D is the incorrect statement, so the answer is D.

1. Recall that carbon can use its $p$-orbitals to form $p\pi-p\pi$ multiple bonds with itself, as seen in double and triple carbon-carbon bonds.

$$\text{C=C}, \quad \text{C}\equiv\text{C}$$

So option A is a correct statement.

2. Recall that boron is small enough to complete its octet in $\text{BCl}_3$ using only three bonds (with some back-bonding), so it stays a monomer, whereas aluminium tends to complete its octet by forming a coordinate bond with another $\text{AlCl}_3$ unit.

$$\text{BCl}_3 \ (\text{monomer}), \quad \text{Al}_2\text{Cl}_6 \ (\text{dimer of AlCl}_3)$$

So option B is a correct statement.

3. Recall that catenation strength in Group 14 decreases down the group as the element-element bond becomes weaker.

$$\text{C} \gg \text{Si} > \text{Ge} \approx \text{Sn}$$

So option C is a correct statement.

4. Check option D: although oxygen commonly shows $-2$ in compounds like water, it also shows other oxidation states in different compounds.

$$\text{O in O}_2 \Rightarrow 0, \quad \text{O in H}_2\text{O}_2 \Rightarrow -1, \quad \text{O in H}_2\text{O} \Rightarrow -2$$

Since oxygen does not show only the $-2$ state, option D is the incorrect statement.

Hence, the answer is D.