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AIPMT 2002 – Periodic Table & Classification of Elements (Periodic Properties Trend)
NEET Q.1 : Which of the following order is wrong?
(a) $NH_3 < PH_3 < AsH_3$ – Acidic strength
(b) $Li < Be < B < C$ – 1st Ionisation potential
(c) $Al_2O_3 < MgO < Na_2O < K_2O$ – Basic character
(d) $Li^+ < Na^+ < K^+ < Cs^+$ – Ionic radius
[CBSE AIPMT 2002]
Answer : Correct Option: (b)
Step 1: General trend of ionisation potential
- Across a period (left to right), the first ionisation potential increases because atomic size decreases and nuclear charge increases.
- Hence, in the second period:
$$Li < Be < B < C$$ would be expected.
Step 2: Exception between Be and B
- Beryllium (Be): configuration = $1s^2 \, 2s^2$ → stable, completely filled $2s$ subshell.
- Boron (B): configuration = $1s^2 \, 2s^2 \, 2p^1$ → less stable, easier to remove the outer electron.
- As a result:
$$I.P.(Be) > I.P.(B)$$
Step 3: Corrected order
The actual first ionisation potential order is:
$$Li < B < Be < C$$
Thus, the given order in option (b) is wrong, which makes it the correct answer to the question.
This exception-based question is a classic example in NEET PYQs Chapterwise Solutions for Periodic Table Properties by Anand Classes.
📝 Concept Takeaway
- Across a period: ionisation potential increases.
- Exception: $Be$ has higher ionisation potential than $B$ due to stable $2s^2$ configuration.
- Correct order: $Li < B < Be < C$.
AIPMT 2001 – Periodic Table & Classification of Elements (Ionization Enthalpy Trend With Exceptions)
NEET Q.2 : Correct order of 1st ionisation potential (IP) among the following elements: Be, B, C, N, O
(a) $B < Be < C < O < N$
(b) $B < Be < C < N < O$
(c) $Be < B < C < N < O$
(d) $Be < B < C < O < N$
[CBSE AIPMT 2001]
Answer : Correct Option: (a)
Step 1: General periodic trend
- Across a period (from left to right), ionisation potential increases because of increasing nuclear charge and decreasing atomic size.
- Thus, we would expect:
$$Be < B < C < N < O$$
Step 2: Exceptions in the trend
- Boron (B): configuration = $1s^2 \, 2s^2 \, 2p^1$ → removal of one $2p$ electron is relatively easier.
- Beryllium (Be): configuration = $1s^2 \, 2s^2$ → stable filled $2s$ subshell, harder to remove an electron.
So, $I.P.(B) < I.P.(Be)$. - Nitrogen (N): configuration = $1s^2 \, 2s^2 \, 2p^3$ → half-filled $p$-subshell, extra stability.
- Oxygen (O): configuration = $1s^2 \, 2s^2 \, 2p^4$ → electron–electron repulsion in paired $p$ orbital makes removal easier.
So, $I.P.(N) > I.P.(O)$.
Step 3: Correct order
Thus, the actual ionisation potential order is:
$$B < Be < C < O < N$$
This matches option (a) and is a frequently tested exception in NEET PYQs Chapterwise MCQS Solutions and Important Problems for Periodic Table Properties Chapter Class 11 Chemistry.
📝 Concept Takeaway
- Across a period: IP increases, but exceptions occur.
- $B < Be$ (because Be has a stable $2s^2$ configuration).
- $O < N$ (because N has a half-filled $2p^3$ subshell, more stable).
- Correct order: $B < Be < C < O < N$.
AIPMT 1998 – Periodic Table & Classification of Elements (Ionization Potential Class 11 Chemistry)
NEET Q.3 : The first ionisation potential (in eV) of Be and B, respectively, are
(a) 8.29, 9.32
(b) 9.32, 9.32
(c) 8.29, 8.29
(d) 9.32, 8.29
[CBSE AIPMT 1998]
Answer : Correct Option: (d)
Step 1: General trend
Across a period (from left to right), ionisation potential generally increases because of increasing nuclear charge. So, one might expect:
$$I.P.(Be) < I.P.(B)$$
Step 2: Exception due to configuration
- Beryllium (Be): electronic configuration = $1s^2 \, 2s^2$
- Stable, fully filled $2s$ subshell.
- Removing an electron from $2s$ is relatively difficult.
- Boron (B): electronic configuration = $1s^2 \, 2s^2 \, 2p^1$
- The outermost electron is in the $2p$ orbital.
- $2p$ orbital is higher in energy and less tightly bound than $2s$.
- Easier to remove compared to Be.
Thus, $I.P.(Be) > I.P.(B)$.
Step 3: Actual values
- Ionisation potential of Be = 9.32 eV
- Ionisation potential of B = 8.29 eV
So, the correct order is:
$$I.P.(Be) = 9.32 \, eV > I.P.(B) = 8.29 \, eV$$
This matches option (d) and illustrates an important exception in NEET PYQs Chapterwise Solutions of Class 11 Chemistry for Periodic Table Properties by Anand Classes.
📝 Concept Takeaway
- Across a period, ionisation potential increases, but exceptions occur.
- Stable filled subshells (Be: $2s^2$) → higher I.P.
- Electrons in $p$ orbital (B: $2p^1$) → lower I.P.
- Values: $I.P.(Be) = 9.32 \, eV$, $I.P.(B) = 8.29 \, eV$.
AIPMT 1998 – Periodic Table & Classification of Elements (Ionic Radii Class 11 Chemistry)
NEET Q.4 : In crystals of which of the following ionic compounds would you expect maximum distance between centres of cations and anions?
(a) LiF
(b) CsF
(c) CsI
(d) LiI
[CBSE AIPMT 1998]
Answer : Correct Option: (c)
Step 1: Concept of ionic distance
The distance between the centres of cations and anions in an ionic crystal depends on the sizes of the ions.
- Larger cation radius → larger separation.
- Larger anion radius → larger separation.
Step 2: Trend in cation size (alkali metals)
Group 1 cations ($Li^+, Na^+, K^+, Rb^+, Cs^+$):
- Ionic radius increases down the group.
- Thus, $Cs^+$ has the largest size among the listed cations.
Step 3: Trend in anion size (halides)
Group 17 anions ($F^-, Cl^-, Br^-, I^-$):
- Ionic radius increases down the group.
- Thus, $I^-$ has the largest size among the listed anions.
Step 4: Comparison of given compounds
| Compound | Cation ($M^+$) | Approx. Radius (pm) | Anion ($X^-$) | Approx. Radius (pm) | Distance (M⁺ + X⁻) |
|---|---|---|---|---|---|
| LiF | $Li^+$ | ~76 | $F^-$ | ~133 | ~209 |
| CsF | $Cs^+$ | ~167 | $F^-$ | ~133 | ~300 |
| LiI | $Li^+$ | ~76 | $I^-$ | ~220 | ~296 |
| CsI | $Cs^+$ | ~167 | $I^-$ | ~220 |
Step 5: Combining effects
The maximum cation–anion distance will be when the largest cation (Cs⁺) pairs with the largest anion (I⁻).
Therefore,
$$\text{Maximum distance in crystal} = \text{CsI}$$
So, the correct option is (c), consistent with trends explained in NEET and AIIMS PYQs Chapterwise Solutions of Class 11 Chemistry for Periodic Table Properties.
📝 Concept Takeaway
- Ionic distance increases with increasing ionic radii of both cation and anion.
- Down the group, cation size ↑ and anion size ↑.
- Maximum separation occurs in $CsI$ because $Cs^+$ and $I^-$ are the largest ions in their respective groups.
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