The unruly crowd demanded that the accused be
(A) hanged
(B) hanging
(C) hankering
without trial.
(D) hung
Choose the statement(s) where the underlined word is used correctly:
(i)
(ii)
(iii)
A prone is a dried plum.
He was lying prone on the floor.
People who eat a lot of fat are prone to heart disease.
(A) (i) and (iii) only (B) (iii) only
(C) (i) and (ii) only (D) (ii) and (iii) only
Options:
(A) Raman active but not NMR (nuclear magnetic resonance) active.
(B) Infrared active and Raman active but not NMR active.
(C) Raman active and NMR active.
(D) Only NMR active.
Official Answer Key:
D
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2016 Q.3)**
**1. Molecular Selection Rules**:
- **Infrared (IR) Activity**: Requires a changing electric dipole moment during vibration ($\frac{d\mu}{dq} \neq 0$). For a homonuclear diatomic molecule like $^{17}\text{O}_2$, the permanent and dynamic dipole moment is identically zero, so it is **IR inactive**.
- **Raman Activity**: Requires a change in molecular polarizability during vibration ($\frac{d\alpha}{dq} \neq 0$). The symmetric stretching vibration of homonuclear diatomics alters the electron cloud size, hence it is **Raman active**.
- **Nuclear Magnetic Resonance (NMR) Activity**: Requires non-zero nuclear spin ($I \neq 0$). The oxygen isotope $^{17}\text{O}$ has an odd mass number ($A = 17, Z = 8$), giving nuclear spin $I = 5/2 \neq 0$. Therefore, $^{17}\text{O}_2$ possesses a nuclear magnetic moment and is **NMR active**.
**2. Conclusion**:
- Combining all spectroscopic modalities: $^{17}\text{O}_2$ is **Raman active and NMR active** (and IR inactive).
**3. Official Verification**:
The verified official IIT Master Answer is **Option (C)**.
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