Paper: CSIR-NET June 2019 · Subject: Inorganic Chemistry · Chapter: Main Group Elements · Topic: Main Group Elements – General · Marks: 2 · Difficulty: Easy
The correct statements regarding Boron among the following (1) Nuclear spin of ${ }^{11} \mathrm{B}$ is greater than that of ${ }^{10} \mathrm{B}$ (II) The polarities of B-H bond and C-H bonds are opposite (III) Cross-section of neutron absorption for ${ }^{10} \mathrm{B}$ is much more than that of ${ }^{11} \mathrm{B}$ (IV) B reacts with boiling aq. NaOH solution to form $\mathrm{NaB}(\mathrm{OH})_{4}$
(a)II and III
(b)I and II
(c)III and IV
(d)II and IV
Answer
Answer: A ✓ checked by 4AB · confidence medium
Explanation
B–H (B^δ+–H^δ−) is polarised opposite to C–H. ¹⁰B has a huge neutron cross-section. ¹¹B (I = 3/2) has a smaller spin than ¹⁰B (I = 3). Crystalline boron is not attacked by boiling aqueous NaOH. II and III (a).