Subject: Inorganic Chemistry · Chapter: Main Group Elements · Topic: Main Group Elements – General · Exam: CSIR-NET 2019 · 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. $\mathrm{NaOH}$ solution to form $\mathrm{NaB(OH)_4}$Are
(a)II and III
(b)I and II
(c)III and IV
(d)II and IV
Answer
Answer (as printed): A
Explanation
• Nuclear spin of $^{10}\mathrm{B} = 3$; nuclear spin of $^{11}\mathrm{B} = 3/2$.$\mathrm{B} \rightarrow \mathrm{H}$; $\mathrm{C} \leftarrow \mathrm{H}$. Electronegativity: 2.04, 2.1, 2.5, 2.1. This indicates that polarities of B–H and C–H bonds are opposite.• Thermal neutron absorption cross section for $^{10}\mathrm{B}=3837$ Barn and for $^{11}\mathrm{B}=0.005$ Barn.• Boron resists attack by boiling concentrated aqueous NaOH or fused NaOH up to $500^{\circ}\mathrm{C}$.$\mathrm{B} + \mathrm{NaOH(aq)} \rightarrow$ No reaction.Boron reacts with fused alkali to give sodium metaborate and $\mathrm{H_2}$.