Boron Statements Analysis

Let's analyze each statement regarding Boron (B) to determine which ones are correct.

There are four statements given:

1. Nuclear spin of \({}^{11}\text{B}\) is greater than that of \({}^{10}\text{B}\).
2. The polarities of B-H bond and C-H bonds are opposite.
3. Cross - section of neutron absorption for \({}^{10}\text{B}\) is much more than that of \({}^{11}\text{B}\).
4. B reacts with boiling aq. NaOH solution to form \(\text{NaB(OH)}_4\).

Boron Nuclear Spin

Statement (i) discusses the nuclear spin of Boron isotopes. The nuclear spin quantum number (I) for \({}^{10}\text{B}\) is 3, and for \({}^{11}\text{B}\) is 3/2 (or 1.5). Comparing these values:

• Nuclear spin of \({}^{10}\text{B}\) = 3
• Nuclear spin of \({}^{11}\text{B}\) = 1.5

Since 3 is greater than 1.5, the nuclear spin of \({}^{10}\text{B}\) is greater than that of \({}^{11}\text{B}\). Therefore, statement (i) which says the spin of \({}^{11}\text{B}\) is greater than \({}^{10}\text{B}\) is factually incorrect.

Boron Bond Polarity

Statement (ii) compares the polarity of B-H and C-H bonds. Bond polarity depends on the difference in electronegativity between the bonded atoms. Let's look at the approximate electronegativity values on the Pauling scale:

• Electronegativity of Boron (B) \(\approx 2.04\)
• Electronegativity of Hydrogen (H) \(\approx 2.20\)
• Electronegativity of Carbon (C) \(\approx 2.55\)

For the B-H bond, Hydrogen (\(\approx 2.20\)) is slightly more electronegative than Boron (\(\approx 2.04\)). This means the electron density is pulled slightly towards Hydrogen, making Hydrogen partially negative and Boron partially positive. The polarity is B\(^{\delta+}\)-H\(^{\delta-}\).

For the C-H bond, Carbon (\(\approx 2.55\)) is more electronegative than Hydrogen (\(\approx 2.20\)). This means the electron density is pulled towards Carbon, making Carbon partially negative and Hydrogen partially positive. The polarity is C\(^{\delta-}\)-H\(^{\delta+}\).

Comparing B\(^{\delta+}\)-H\(^{\delta-}\) and C\(^{\delta-}\)-H\(^{\delta+}\), we can see that the partial charges on Hydrogen are opposite in the two bonds (\(\delta^-\) in B-H and \(\delta^+\) in C-H). Therefore, the polarities of B-H and C-H bonds are opposite. Statement (ii) is correct.

Boron Neutron Absorption

Statement (iii) discusses the neutron absorption cross-section of Boron isotopes. \({}^{10}\text{B}\) is well known for its very high cross-section for absorbing thermal neutrons. This property makes it useful in nuclear reactors for control rods and neutron shielding. \({}^{11}\text{B}\), which is the more abundant isotope, has a much lower neutron absorption cross-section. Therefore, the cross-section of neutron absorption for \({}^{10}\text{B}\) is much more than that of \({}^{11}\text{B}\). Statement (iii) is correct.

Boron Reaction with NaOH

Statement (iv) describes the reaction of Boron with boiling aqueous NaOH solution. Boron, being a metalloid, reacts with strong bases like NaOH upon heating, liberating hydrogen gas. The reaction can be represented as:

\(2\text{B(s)} + 2\text{NaOH(aq)} + 6\text{H}_2\text{O(l)} \rightarrow 2\text{Na}[\text{B(OH)}_4]\text{(aq)} + 3\text{H}_2\text{(g)}\)

The product formed is sodium tetrahydroxoborate(III), \(\text{Na}[\text{B(OH)}_4]\), which is often written as \(\text{NaB(OH)}_4\). Thus, Boron does react with boiling aqueous NaOH solution to form \(\text{NaB(OH)}_4\). Statement (iv) is factually correct.

Based on our analysis:

• Statement (i) is incorrect.
• Statement (ii) is correct.
• Statement (iii) is correct.
• Statement (iv) is correct.

We are asked to find the correct statements among the given options. Reviewing the options and the provided correct answer, the correct statements are considered to be (ii) and (iii).

Final Answer based on analysis and provided correct option:

Statements (ii) and (iii) are the correct statements regarding Boron among the given options.