Question

How many of the following trend is "INCORRECT" ?

• A. Electron affinity : Se > S > Br > Cl
• B. Electronegativity : B > Al > Ga > In
• C. Atomic radii : S > Cl > O > F
• D. $IE_2(Be) > I.E_2(Li) > IE_1(Li)$

Answer 1

3

Answer 2

The question asks us to identify the number of incorrect trends among the given options. Let's analyze each option:

(A) Electron affinity : Se > S > Br > Cl

• Electron affinity (EA) generally increases across a period and decreases down a group.

• There's an anomaly: Second-period elements (like O, F) generally have lower electron affinities than their third-period counterparts (S, Cl) due to their small size and increased interelectronic repulsion.

• Comparing within a group (downwards): EA decreases.

  • Group 16: S > Se
  • Group 17: Cl > Br

• Comparing within a period (left to right): EA generally increases.

  • Period 3: Cl > S
  • Period 4: Br > Se

• Combining these facts:

  • Cl has the highest electron affinity among halogens and generally among non-metals.
  • Comparing Cl and S (both Period 3): Cl > S.
  • Comparing Br and Se (both Period 4): Br > Se.
  • Comparing S and Br: S is in Period 3, Br in Period 4. Cl > S and Cl > Br. Br is more electronegative than S, so Br has higher electron affinity than S. (Actual values: Cl (-349 kJ/mol) > Br (-325 kJ/mol) > S (-200 kJ/mol) > Se (-195 kJ/mol)).

• Therefore, the correct order is: Cl > Br > S > Se.

• The given trend Se > S > Br > Cl is INCORRECT.

(B) Electronegativity : B > Al > Ga > In

• Electronegativity (EN) generally decreases down a group.

• For Group 13 elements (B, Al, Ga, In, Tl), the expected trend would be B > Al > Ga > In > Tl.

• However, due to the poor shielding effect of d-electrons (in Ga) and d & f-electrons (in Tl), the effective nuclear charge for Ga and Tl is higher than expected. This causes an anomalous trend in electronegativity.

  • Ga has higher EN than Al (Ga > Al) because of the poor shielding by 3d electrons in Ga.
  • Tl has higher EN than In (Tl > In) because of the poor shielding by 4f and 5d electrons in Tl.

• The actual trend for Group 13 electronegativity is approximately: B > Tl $\approx$ Ga $\approx$ In > Al.

• Since Al < Ga, the given trend B > Al > Ga > In is INCORRECT.

(C) Atomic radii : S > Cl > O > F

• Atomic radii generally decrease across a period (from left to right) and increase down a group (from top to bottom).

• Within Period 2: Oxygen (O) is to the left of Fluorine (F). So, O > F.

• Within Period 3: Sulfur (S) is to the left of Chlorine (Cl). So, S > Cl.

• Comparing across periods (down a group):

  • Sulfur (S) is below Oxygen (O). So, S > O.
  • Chlorine (Cl) is below Fluorine (F). So, Cl > F.

• Combining these: S is the largest, followed by Cl (smaller than S, but larger than O and F). O is larger than F.

• Therefore, the correct order is: S > Cl > O > F.

• The given trend S > Cl > O > F is CORRECT.

(D) $IE_2(Be) > I.E_2(Li) > IE_1(Li)$

• Ionization Energy (IE) is the energy required to remove an electron.
• $IE_1(Li)$: Removal of the first electron from Li ($1s^2 2s^1 \rightarrow 1s^2$). This is the removal of a 2s electron.
• $IE_2(Li)$: Removal of the second electron from $Li^+$ ($1s^2 \rightarrow 1s^1$). This involves removing an electron from a completely filled inner 1s shell (noble gas configuration). This requires a very large amount of energy.
• $IE_2(Be)$: Removal of the second electron from $Be^+$ ($1s^2 2s^1 \rightarrow 1s^2$). This involves removing a 2s electron from a cation.
• Comparing $IE_2(Li)$ and $IE_2(Be)$: $IE_2(Li)$ involves removing an electron from a stable $1s^2$ configuration, which is much harder than removing a 2s electron from $Be^+$. So, $IE_2(Li)$ must be significantly greater than $IE_2(Be)$.
• Comparing $IE_2(Be)$ and $IE_1(Li)$: Both involve removing a 2s electron. However, for $IE_2(Be)$, the electron is removed from $Be^+$ (a cation with Z=4), while for $IE_1(Li)$, it's removed from neutral Li (Z=3). The effective nuclear charge on the 2s electron in $Be^+$ is much higher than in Li. Thus, $IE_2(Be) > IE_1(Li)$.
• Therefore, the correct order is: $IE_2(Li) > IE_2(Be) > IE_1(Li)$.
• The given trend $IE_2(Be) > I.E_2(Li) > IE_1(Li)$ is INCORRECT.

Out of the four trends, (A), (B), and (D) are incorrect. So, there are 3 incorrect trends.