Question

which element will form the ion with the smallest ionic radius? periodic table image with h, he, li, be, b, c, n, o, f, ne, na, mg, al, si, p, s, cl, ar, k, ca, ga, ge, as, se, br, kr, rb, sr, in, sn, sb, te, i, xe, cs, ba, tl, pb, bi, po, at, rn, fr, ra, nh, fl, mc, lv, ts, og. options: be, li, b

Explanation:

Step1: Analyze ion formation

Li forms \( \text{Li}^+ \), Be forms \( \text{Be}^{2+} \), B forms \( \text{B}^{3+} \). These are all cations (lost electrons), so they have the same electron configuration as the nearest noble gas (He for Li⁺, Be²⁺, B³⁺).

Step2: Relate ionic radius to nuclear charge

For isoelectronic species (same electron configuration), ionic radius decreases with increasing nuclear charge (number of protons). Li has 3 protons, Be has 4 protons, B has 5 protons.

Step3: Compare nuclear charges

Since \( \text{B}^{3+} \) has the highest nuclear charge (5 protons) among \( \text{Li}^+ \) (3), \( \text{Be}^{2+} \) (4), and \( \text{B}^{3+} \), it will pull the electrons most strongly, resulting in the smallest ionic radius. Wait, but wait—wait, let's check again. Wait, Li⁺: 3 protons, Be²⁺: 4 protons, B³⁺: 5 protons. Same electron cloud (He configuration). So higher nuclear charge → smaller radius. But wait, the options are Be, Li, B. Wait, no—wait, the question is which element (Be, Li, B) forms the ion with smallest ionic radius. Wait, Li⁺: 1s², Be²⁺: 1s², B³⁺: 1s². So same electron configuration. Nuclear charge: Li (3), Be (4), B (5). So B³⁺ has the smallest radius? But wait, the options are Be, Li, B. Wait, maybe I made a mistake. Wait, no—wait, let's check the elements. Wait, Li is in group 1, Be group 2, B group 13. When they form cations: Li loses 1 e⁻ (Li⁺), Be loses 2 e⁻ (Be²⁺), B loses 3 e⁻ (B³⁺). All have electron configuration of He (2 electrons). So the number of protons: Li (3), Be (4), B (5). The more protons, the stronger the attraction to the electrons, so the smaller the radius. So B³⁺ should be smallest. But wait, the options are Be, Li, B. Wait, the question's options are the elements: Be, Li, B. Wait, maybe I misread. Wait, the problem is asking which element (Be, Li, B) forms the ion with smallest ionic radius. So the ions are Li⁺, Be²⁺, B³⁺. Since B³⁺ has the highest nuclear charge (5 protons) among them, it has the smallest ionic radius. But wait, the options are Be, Li, B. Wait, the answer should be B? But wait, let's check again. Wait, Li⁺: radius ~0.76 Å, Be²⁺: ~0.45 Å, B³⁺: ~0.23 Å. So yes, B³⁺ is smaller. But wait, the options are Be, Li, B. So the element is B? Wait, no—wait, the options are the elements: Be, Li, B. So the answer is B? Wait, but maybe I made a mistake. Wait, let's re-express: for isoelectronic cations, ionic radius decreases with increasing positive charge (which is related to nuclear charge). Li⁺ (charge +1, protons 3), Be²⁺ (charge +2, protons 4), B³⁺ (charge +3, protons 5). So B³⁺ has the smallest radius. So the element is B. But wait, the options are Be, Li, B. So the answer is B? Wait, but maybe the question is about the ions formed by these elements, and we have to choose the element whose ion has smallest radius. So B's ion (B³⁺) has smallest radius. So the answer is B.

Answer:

B