NEET 2022 Chemistry Hydrides Match the Following Question

Chapter: Chemical Bonding and Molecular Structure

Class: 11 | Topic: Classification of Hydrides | Difficulty: 🟡 Moderate

✅ Ans: (A) — (a)-iv, (b)-i, (c)-ii, (d)-iii

• $MgH_2$ → Ionic (iv) Mg is an electropositive metal and forms $Mg^{2+}$ and $H^-$. Hence, it is an ionic hydride.

• $GeH_4$ → Electron precise (i) Ge has 4 valence electrons and forms four $Ge-H$ covalent bonds. It has the required number of electrons for normal covalent bonding → electron precise.

• $B_2H_6$ → Electron deficient (ii) Diborane has only 12 valence electrons and cannot form all normal 2-centre–2-electron bonds. It contains two 3-centre–2-electron bonds → electron deficient.

• $HF$ → Electron rich (iii) HF has a highly electronegative F atom with three lone pairs, making it electron rich.

❌ Why other options are wrong?

They interchange the characteristic nature of these hydrides. The key distinctions are:

$$\boxed{MgH_2=\text{Ionic}}$$ $$\boxed{GeH_4=\text{Electron precise}}$$ $$\boxed{B_2H_6=\text{Electron deficient}}$$ $$\boxed{HF=\text{Electron rich}}$$

📌 NCERT: Hydrides are classified based on the nature of the hydrogen–element bond; electron-deficient hydrides such as $B_2H_6$ have fewer electrons than required for conventional two-centre bonds.

🧠 NEET Trick: Mg → Ionic | Ge → Precise | B → Deficient | F → Rich.