Answer: [Fe(H 2 O) 5 NO] 2+.
- A [Fe(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup>
- B [Fe(H<sub>2</sub>O)<sub>5</sub>NO]<sup>2+</sup>
- C [Fe(NO)<sub>2</sub>]<sup>2+</sup>
- D [Fe(H<sub>2</sub>O)<sub>5</sub>(OH)]<sup>2+</sup>
Correct answer: B. [Fe(H<sub>2</sub>O)<sub>5</sub>NO]<sup>2+</sup>
Explanation: In the brown ring test, NO reduced from the nitrate combines with Fe<sup>2+</sup> to form the brown complex [Fe(H<sub>2</sub>O)<sub>5</sub>NO]<sup>2+</sup> in which iron is in the +1 state.
Xenon tetrafluoride (XeF4) has six electron domains around xenon — four Xe–F bonds plus two lone pairs. The lone pairs occupy opposite axial positions, leaving the four fluorine atoms in a square-planar shape (Xe–F ≈ 194 pm). Image: ChemSim, Public Domain, via Wikimedia Commons.
Concept context
Covers the nitrogen family (Group 15), oxygen family (Group 16), halogens (Group 17), and noble gases (Group 18). One of the highest-weightage inorganic chapters in NEET and JEE - includes preparation and properties of HNO₃, H₂SO₄, interhalogens, and xenon compounds.