Question

Identify the incorrect statement from the following:

• A. \(ECl_3\), \((E=B \text{ and } Al)\), is a monomer when \(E=B\) and a dimer when \(E=Al\).
• B. The order of catenation property of Group 14 elements is \(C >> Si>Ge = Sn\).
• C. Oxygen exhibits only \(-2\) oxidation state.
• D. Carbon has the ability to form \(p\pi-p\pi\) multiple bond with itself.

Hint:

Oxygen commonly shows \(-2\), but it can also show \(-1\), \(-\frac{1}{2}\), \(+1\), and \(+2\) oxidation states in special compounds.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question requires an evaluation of various descriptive chemistry facts concerning Group 13, 14, and 16 elements. We must identify which statement is scientifically false by examining bonding behaviors ($p\pi-p\pi$), structural tendencies (monomer vs dimer), periodic trends (catenation), and oxidation states.
Step 2: Key Formula or Approach:
The approach is to use the first-principles knowledge of inorganic chemistry:
Second-period elements ($C, N, O$) can form multiple bonds via $p\pi-p\pi$ overlap.
Lewis acidity in Group 13 determines whether compounds exist as monomers or dimers.
Catenation is the ability of an element to form long chains and depends on bond dissociation energy.
Oxidation states of highly electronegative elements like Oxygen depend on the partner atom's electronegativity.
Step 3: Detailed Explanation:

Statement A: Carbon is small and has the correct orbital size to form effective lateral overlaps. Thus, it forms stable double and triple bonds ($C=C, C \equiv C$). This is correct.
Statement B: $BCl_3$ is a monomer because boron is small and can satisfy its octet partially via back-bonding from chlorine. $AlCl_3$ is larger and forms a stable dimer $Al_2Cl_6$ through chlorine bridge bonding to complete the aluminum octets. This is correct.
Statement C: Catenation depends on the $M-M$ bond strength. $C-C$ is the strongest ($348 kJ/mol$), while $Si-Si$, $Ge-Ge$, and $Sn-Sn$ are significantly weaker. The order $C \gg Si>Ge \approx Sn$ is the standard trend. This is correct.
Statement D: While $-2$ is common, Oxygen shows $-1$ in peroxides ($H_2O_2$), $-1/2$ in superoxides ($KO_2$), and even positive oxidation states ($+1, +2$) when bonded to Fluorine (e.g., $O_2F_2$ and $OF_2$). Therefore, saying it shows "only" $-2$ is incorrect.
Step 4: Final Answer:
Statement (D) is the incorrect one.