Q.Metallic radii of some transition elements are given below. Which of these elements will have highest density? Element: Fe, Co, Ni, Cu
Metallic radii/pm: Fe $= 126$, Co $= 125$, Ni $= 125$, Cu $= 128$

Concept & Intuition

Ionisation enthalpy jumps sharply when you try to remove an electron from a stable, half-filled or fully-filled subshell. For transition metals, the third ionisation often involves removing an electron from the 3d subshell. If the +2 ion already has a half-filled 3d⁵ configuration, removing a third electron destroys that stability, requiring a lot of energy. So we look for the element whose +2 ion has exactly 3d⁵.

Step-by-step reasoning

1. Write the ground-state electronic configurations (using the Aufbau principle, remembering that 4s fills before 3d but empties first in ionisation):

   • V (Z=23): [Ar] 3d³ 4s²
   • Cr (Z=24): [Ar] 3d⁵ 4s¹ (special stability of half-filled 3d)
   • Mn (Z=25): [Ar] 3d⁵ 4s²
   • Fe (Z=26): [Ar] 3d⁶ 4s²

2. First two ionisations remove the 4s electrons (and possibly one 3d for Cr, because its 4s¹ is easily lost):

   • V → V²⁺: [Ar] 3d³
   • Cr → Cr²⁺: [Ar] 3d⁴ (loses the 4s¹ and one 3d)
   • Mn → Mn²⁺: [Ar] 3d⁵ (half-filled, very stable)
   • Fe → Fe²⁺: [Ar] 3d⁶

3. Third ionisation removes one more electron from the 3d subshell:

   • V³⁺: 3d² (removing from 3d³ — no special stability lost)
   • Cr³⁺: 3d³ (removing from 3d⁴ — no special stability lost)
   • Mn³⁺: 3d⁴ (removing from 3d⁵ — destroys the half-filled stability)
   • Fe³⁺: 3d⁵ (removing from 3d⁶ — actually gains half-filled stability, so this ionisation is lower than expected)

4. Compare the energy cost:

   The half-filled 3d⁵ configuration in Mn²⁺ is exceptionally stable. To remove an electron from it requires a large amount of energy — the third ionisation enthalpy of Mn is the highest among these four.

   In contrast, Fe²⁺ → Fe³⁺ actually creates a half-filled 3d⁵, so its third ionisation enthalpy is lower than Mn’s.

Step 1 — The concept: why lanthanoids are normally +3

The characteristic oxidation state of the entire lanthanoid series is +3, formed by losing the two $6s$ electrons and one $4f$ (or $5d$) electron. Any state other than +3 appears only when there is a special stability to be gained, namely an

• empty f-subshell, $4f^{0}$,
• half-filled f-subshell, $4f^{7}$, or
• completely filled f-subshell, $4f^{14}$.

This single rule explains every "anomalous" lanthanoid oxidation state you are asked to remember.

Step 2 — Apply the rule to Cerium (Z = 58)

Ground-state configuration:

Removing all four of those outer electrons ($1 + 1 + 2 = 4$) gives:

The ion is left with the complete, noble-gas xenon core — an $f^0$ configuration. That is a strong thermodynamic incentive, so Ce readily shows +4. In practice $\ce{Ce^{4+}}$ salts (ceric ammonium nitrate/sulphate) are stable, commercially used oxidising agents:

(The large positive $E^\circ$ tells us $\ce{Ce^{4+}}$ does still want to revert to the +3 state — it is a good oxidant — but it is kinetically stable enough to bottle and use, which is what "well known to exhibit +4" means.)

Step 3 — Why the other three fail

Note the trap: Samarium and Europium are famous for an anomalous state — but it is +2, not +4. (The other lanthanoid that shows +4 is Terbium, $\ce{Tb^{4+}} = 4f^7$, half-filled — but it is not among the options.)

ANSWER: D

The key concept here is periodic trends in ionisation enthalpy and the anomalous drop at Group 13. Normally, ionisation enthalpy increases across a period (left to right) as nuclear charge increases and atomic radius decreases. However, there is a well-known dip at aluminium (Group 13) compared to magnesium (Group 2) because aluminium’s outermost electron is in a 3p orbital, which is slightly higher in energy and more shielded than the 3s orbital of magnesium. This makes it easier to remove.

We are given:

• Na (Group 1): 497 kJ mol⁻¹
• Mg (Group 2): 738 kJ mol⁻¹
• Si (Group 14): 787 kJ mol⁻¹

We need Al (Group 13). Let’s reason step by step.

1. Observe the trend from Na to Mg to Si

   From Na to Mg, the increase is $738 - 497 = 241$ kJ mol⁻¹.

   From Mg to Si, the increase is $787 - 738 = 49$ kJ mol⁻¹.

   The jump from Na to Mg is large because Mg has a full 3s subshell and higher nuclear charge. The jump from Mg to Si is smaller because Si’s electron is in the same shell but with higher nuclear charge.

2. Estimate where Al should lie

   If the trend were perfectly linear across the period, Al (between Mg and Si) would have an ionisation enthalpy roughly halfway between 738 and 787:

But we know this is not the case — Al has a lower value than this linear estimate because of the s-to-p orbital change.

3. Account for the drop due to the p-orbital electron

   The drop from Mg to Al is a classic exception. Typically, the first ionisation enthalpy of Al is about 10–15% lower than that of Mg. A common empirical observation is that Al’s value is close to that of Na plus a small increment, but here we can use the given data more directly.

   Notice that the increase from Na to Mg is 241, while from Mg to Si is only 49. The drop at Al means its value should be less than Mg’s value of 738. How much less? A reasonable estimate: the difference between Mg and Al is often around 150–160 kJ mol⁻¹.

This matches option (B).

4. Check consistency with Si

   If Al were 578, then the increase from Al to Si is $787 - 578 = 209$ kJ mol⁻¹, which is plausible because Si has a higher nuclear charge and a p-orbital electron that is less shielded.

5. Eliminate other options

   • (A) 769 is too close to Mg and Si — no drop.
   • (C) 488 is too low (even lower than Na).
   • (D) 857 is higher than Si — impossible for Al.

Step 1 — The spin-only formula.

For a transition-metal ion whose magnetism comes from electron spin alone (orbital contribution quenched, as is usual for first-row ions):

where $n$ is the number of unpaired electrons. So a measured $\mu$ can be inverted to count the unpaired electrons — that is the whole strategy here.

Step 2 — Solve for $n$.

A count of electrons cannot be negative, so

(Quick check: $\mu = \sqrt{4 \times 6} = \sqrt{24} = 4.90$ BM ✓.)

Step 3 — Write the configurations of the candidate iron ions.

Iron has $Z = 26$: $\mathrm{Fe} = [\mathrm{Ar}]\,3d^6 4s^2$. Remove the $4s$ electrons first, then $3d$. Fill the five $d$ orbitals by Hund's rule (singly first, then pair up):

Step 4 — Read off the answer.

Only $\mathrm{Fe^{2+}}$ ($3d^6$) has exactly four unpaired electrons: the sixth $d$ electron is forced to pair up in the first orbital, leaving four orbitals singly occupied. Hence

Note the useful contrast with $\mathrm{Fe^{3+}}$ ($3d^5$, half-filled, $n = 5$, $\mu = \sqrt{35} = 5.92$ BM) — the two common iron ions are cleanly distinguished by their magnetic moments, which is exactly what makes this measurement useful in practice.

ANSWER: C

Step 1 — Check (A): density, $\mathrm{I>Br>Cl>F}$.

Down group 17 the atomic mass rises much faster than the atomic volume, so density increases down the group ($\mathrm{F_2}$ and $\mathrm{Cl_2}$ are gases, $\mathrm{Br_2}$ a liquid, $\mathrm{I_2}$ a solid). The order $\mathrm{I>Br>Cl>F}$ is correct.

Step 2 — Check (C): ionisation enthalpy, $\mathrm{F>Cl>Br>I}$.

Ionisation enthalpy decreases down a group: the outermost electron sits in a shell of larger $n$, further from the nucleus and better shielded, so it is easier to pull out. $\mathrm{F>Cl>Br>I}$ is correct (halogens have very high IE anyway — they want to gain, not lose, an electron).

Step 3 — Check (D): electronegativity, $\mathrm{F>Cl>Br>I}$.

Fluorine is the most electronegative element in the periodic table (4.0 on the Pauling scale), and electronegativity falls down the group as size grows: F (4.0) > Cl (3.2) > Br (3.0) > I (2.7). Correct.

Step 4 — Check (B): electron gain enthalpy — the exception.

Halogens have the most negative $\Delta_{eg}H$ of all groups (one electron completes the octet). But the value is not most negative for fluorine. The observed order of magnitude is

Why? Fluorine's valence shell is the very compact 2p subshell. Squeezing an extra electron into that small volume creates unusually strong electron–electron repulsion, which cancels much of the energy released by the nuclear attraction. Chlorine's 3p subshell is roomier, so the incoming electron feels less repulsion and more net stabilisation. Hence Cl, not F, tops the list. From Cl onwards the normal trend (decreasing down the group, because of increasing size) resumes.

Therefore the match "$\mathrm{F>Cl>Br>I}$ — electron gain enthalpy" is the incorrect one, which is exactly what the question asks for.

ANSWER: B

1. Why d-electron count decides the colour

In an octahedral aqua complex $[\mathrm{M(H_2O)_6}]^{n+}$ the five degenerate d orbitals split into a lower $t_{2g}$ set and an upper $e_g$ set, separated by the crystal-field splitting energy $\Delta_o$. Absorption of a visible photon promotes an electron $t_{2g}\rightarrow e_g$ (a d–d transition), and the ion appears in the complementary colour of the light absorbed.

Such a transition needs (i) at least one electron in the d sub-shell and (ii) at least one vacancy in it. Hence:

2. Configurations of the ions offered

Use $Z$ and remove the 4s electrons first (4s is lost before 3d on ionisation):

3. Test each pair

1. (A) $\mathrm{Sc^{3+}},\ \mathrm{Mn^{2+}}$ — $\mathrm{Sc^{3+}}$ is $d^0$ ⇒ colourless. Pair fails.
2. (B) $\mathrm{Ni^{2+}},\ \mathrm{Ti^{4+}}$ — $\mathrm{Ti^{4+}}$ is $d^0$ ⇒ colourless. Pair fails.
3. (C) $\mathrm{Ti^{3+}},\ \mathrm{Cu^{+}}$ — $\mathrm{Cu^{+}}$ is $d^{10}$ (fully filled, no vacancy) ⇒ colourless. Pair fails.
4. (D) $\mathrm{Mn^{2+}},\ \mathrm{Ti^{3+}}$ — $d^5$ and $d^1$: both partially filled ⇒ both coloured. ✓

(The $\mathrm{Mn^{2+}}$ colour is very faint because its $d^5$ high-spin transitions are spin-forbidden as well as Laporte-forbidden — but it is still a coloured ion, which is what the question asks.)

ANSWER: D

[!TLDR]

Down Group 1 (Li → Cs) the outer electron is farther out and more shielded, so IE1 falls steadily; the element with minimum IE1 among the choices is Cs.

Concept

For the alkali metals (a standard CBSE/NCERT Class 11 periodic-trends idea), moving down a group increases the principal quantum number of the valence electron. The atomic radius grows and inner-shell shielding rises faster than the nuclear charge is felt, so the single ns¹ electron is removed more easily — first ionization enthalpy decreases: Li > Na > K > Rb > Cs.

Solution

The curve is plotted as IE1 versus atomic number Z. The highest point P corresponds to the smallest (top) Group-1 element and the value falls as Z increases, levelling off at the bottom for the largest atoms (points S and T). Among the given actual elements $\mathrm{Na}, \mathrm{K}, \mathrm{Rb}, \mathrm{Cs}$, the order of IE1 is

Hence the element with the minimum IE1 is the heaviest one shown, caesium, matching the lowest point on the levelled-off part of the graph.

[!ANSWER]

(D) Cs has the minimum first ionization enthalpy.

The key is to track the change in oxidation number of chromium. In $\text{Cr}_2\text{O}_7^{2-}$, each Cr is in the +6 oxidation state (since oxygen is -2, and the overall charge is -2: $2x + 7(-2) = -2$ gives $x = +6$). In $\text{Cr}^{3+}$, the oxidation state is +3. So each chromium atom gains 3 electrons during reduction.

Since one $\text{Cr}_2\text{O}_7^{2-}$ ion contains two chromium atoms, the total electrons required per ion is $2 \times 3 = 6$.

Now we scale to the given amount:

1. Find electrons per mole of dichromate

   For 1 mole of $\text{Cr}_2\text{O}_7^{2-}$, the reduction to $\text{Cr}^{3+}$ consumes 6 moles of electrons.

2. Scale to 0.2 mole

   Moles of electrons needed = $0.2 \times 6 = 1.2$.

Step 1 — The concept: metallic conduction needs mobile electrons.

A solid conducts like a metal when its electrons occupy a partially filled band (or when a filled band overlaps an empty one). Electrons can then move freely under an applied field, and — the metallic signature — conductivity decreases as temperature rises (lattice vibrations scatter the electrons).

Step 2 — Test each oxide.

• (A) $\mathrm{SiO_2}$ — a giant covalent (network) solid. Every valence electron is locked in a localised $\mathrm{Si-O}$ $\sigma$ bond, and the band gap is very large. Quartz is a classic insulator.
• (B) $\mathrm{MgO}$ — an ionic solid with the closed-shell ions $\mathrm{Mg^{2+}}$ ($[\mathrm{Ne}]$) and $\mathrm{O^{2-}}$ ($[\mathrm{Ne}]$). The ions are fixed in the lattice and no electrons are free, so solid MgO is an insulator (it conducts only when molten/aqueous, and then ionically, not metallically).
• (C) $\mathrm{SO_2}$ (s) — a molecular solid of discrete $\mathrm{SO_2}$ molecules held by weak van der Waals forces. No free electrons or ions: an insulator.
• (D) $\mathrm{CrO_2}$ — chromium is in the $+4$ state, i.e. $\mathrm{Cr}$ is $d^2$. These $d$ electrons are delocalised through Cr–O–Cr overlap into a partially filled band, so $\mathrm{CrO_2}$ shows metallic conductivity (and is ferromagnetic — it was the magnetic material of choice for high-quality audio tapes). It is the standard NCERT example of an oxide that behaves electrically like a metal, alongside $\mathrm{ReO_3}$, $\mathrm{TiO}$ and $\mathrm{VO}$.

ANSWER: D

Step 1 — Oxidation state of Mn in $\mathrm{MnO_4^-}$.

Oxygen is $-2$; the overall charge is $-1$:

Step 2 — Oxidation state of Mn in $\mathrm{MnO_2}$.

Step 3 — Electrons transferred.

Going from $+7$ to $+4$ is a gain of $3$ electrons per Mn atom (reduction):

(The half-equation balances: charge on the left $= -1 + 4 - 3 = 0$, and on the right $= 0$. ✓)

Step 4 — Convert to charge (Faraday's first law).

One mole of electrons carries one faraday:

Reducing 1 mole of $\mathrm{MnO_4^-}$ needs 3 moles of electrons, hence

Common mistake: answering 5 F — that is the electron change for $\mathrm{MnO_4^-} \rightarrow \mathrm{Mn^{2+}}$ (in acidic medium, $+7 \to +2$), a different reduction. Always compute the change for the product actually named.

ANSWER: B

Step 1 — What bauxite is.

Bauxite is the principal ore of aluminium, $\mathrm{Al_2O_3 \cdot xH_2O}$. Freshly mined bauxite is reddish-brown, not white — the colour itself is the clue to the impurity.

Step 2 — Which impurities are actually present.

The standard impurities are $\mathrm{Fe_2O_3}$ (the major one, responsible for the red colour), $\mathrm{SiO_2}$ and $\mathrm{TiO_2}$. Copper, zinc and chromium oxides are not associated with bauxite deposits.

Step 3 — Why the Bayer process works (the chemistry behind the answer).

Alumina is amphoteric, so hot concentrated NaOH dissolves it:

Ferric oxide is purely basic and does not react with alkali, so it is filtered off as the insoluble residue called red mud. The whole leaching step is designed precisely to remove $\mathrm{Fe_2O_3}$ — which confirms it is the common impurity.

Step 4 — Recovery of pure alumina.

followed by calcination at 1470 K to give pure $\mathrm{Al_2O_3}$.

ANSWER: C