D- and f-Block Elements NEET 2008: Second Ionization Order

Which element has the highest second ionization enthalpy in the 2008 question?

Chromium has the highest second ionization enthalpy, so option A is correct in this d- and f-Block Elements NEET 2008 question. The archive entry is Chemistry, 2008, Slot C, chapter d- and f-Block Elements.

Arrange titanium (atomic number 22), vanadium (23), chromium (24) and manganese (25) from greatest to least second ionization enthalpy. Choose from:

The question bank tags this question hard and assigns a 90-second target. These are bank labels, not measured student performance.

Why must second ionization start from the singly charged ion?

Second ionization enthalpy is the enthalpy required to remove one electron from each ion in one mole of singly charged gaseous ions. It does not mean removing two electrons together:

M^{+} (g) \to M^{2 +} (g) + e^{-}

The neutral atom is not the starting species for this step. First write its configuration, then remove one 4s electron to obtain the singly charged ion:

Titanium:

T i : [A r] 3 d^{2} 4 s^{2} ⟶ T i^{+} : [A r] 3 d^{2} 4 s^{1}

Vanadium:

V : [A r] 3 d^{3} 4 s^{2} ⟶ V^{+} : [A r] 3 d^{3} 4 s^{1}

Chromium:

C r : [A r] 3 d^{5} 4 s^{1} ⟶ C r^{+} : [A r] 3 d^{5}

Manganese:

M n : [A r] 3 d^{5} 4 s^{2} ⟶ M n^{+} : [A r] 3 d^{5} 4 s^{1}

The singly charged chromium ion already has no 4s electron left. The other three singly charged ions still have one, which determines the electron removed next.

Which electron does each ion lose during second ionization?

Titanium, vanadium and manganese lose their remaining 4s electron. Chromium must lose a 3d electron because its 4s subshell is already empty.

Titanium, electron source: 4s.

\underset{starting ion}{\underset{⏟}{T i^{+} : [A r] 3 d^{2} 4 s^{1}}} ⟶ \underset{product}{\underset{⏟}{T i^{2 +} : [A r] 3 d^{2}}} + e^{-}

Vanadium, electron source: 4s.

V^{+} : [A r] 3 d^{3} 4 s^{1} ⟶ V^{2 +} : [A r] 3 d^{3} + e^{-}

Chromium, electron source: 3d.

C r^{+} : [A r] 3 d^{5} ⟶ C r^{2 +} : [A r] 3 d^{4} + e^{-}

Manganese, electron source: 4s.

M n^{+} : [A r] 3 d^{5} 4 s^{1} ⟶ M n^{2 +} : [A r] 3 d^{5} + e^{-}

The decisive contrast is breaking versus retaining a half-filled subshell. Chromium loses an electron from its half-filled 3d subshell; manganese retains its half-filled 3d subshell while losing the remaining 4s electron.

How do half-filled stability and nuclear charge give the full order?

Chromium ranks highest because its second ionization breaks a stable half-filled 3d configuration. Nuclear-charge comparisons then place manganese above vanadium and titanium, and vanadium above titanium.

For chromium, removing one of the starting ion’s five 3d electrons destroys the half-filled arrangement:

[A r] 3 d^{5} \to [A r] 3 d^{4} + e^{-}

Manganese instead loses a 4s electron and forms a stable doubly charged ion:

[A r] 3 d^{5} 4 s^{1} \to [A r] 3 d^{5} + e^{-}

This places manganese below chromium. Forming a stable product does not make manganese the lowest in the set. The attraction holding the electron also matters.

As the official solution explains, manganese’s higher nuclear charge places its second ionization enthalpy above those of vanadium and titanium. For vanadium versus titanium, both ions lose their remaining 4s electron; the higher nuclear charge places vanadium above titanium.

These comparisons apply to the four species given here. Neither half-filled stability nor nuclear charge alone gives an exception-free ranking across the transition series.

What is the correct order, and how can you check it?

Option A follows from these three comparisons:

C r > M n, M n > V, V > T i

The requested order is decreasing, so greatest enthalpy goes on the left:

C r > M n > V > T i

Use this checking routine for related questions:

Identify the ionization number.
Write the starting ion.
Identify the electron removed.
Compare the configurations and nuclear charge.

How can an electron-repulsion-only argument lead to option D?

A possible faulty argument is that each successive element from titanium to manganese has more electrons, so greater repulsion must make the second electron progressively easier to remove. This gives exactly option D: $T i > V > C r > M n$

The method fails because it counts added electrons without considering the accompanying increase in nuclear charge. It also never identifies the actual singly charged ion or the electron being removed.

The singly charged chromium ion provides the direct countercheck. Its second ionization removes a 3d electron from a half-filled subshell, unlike the 4s removal in the other three cases.

Which three related questions test the same method?

Test the method by identifying the electron removed, counting unpaired electrons and changing the ionization number. These are original related practice questions, not additional verified PYQs.

Which of the four singly charged ions loses a 3d electron?

Among singly charged titanium, vanadium, chromium and manganese, chromium does. Its 4s subshell is empty, so second ionization changes:

C r^{+} : [A r] 3 d^{5} \to C r^{2 +} : [A r] 3 d^{4} + e^{-}

The other three singly charged ions each retain one removable 4s electron. Chromium is therefore the only 3d-removal case.

How many unpaired electrons does a free doubly charged vanadium ion have?

It has three unpaired electrons. Remove both 4s electrons from neutral vanadium:

V : [A r] 3 d^{3} 4 s^{2} \to V^{2 +} : [A r] 3 d^{3} + 2 e^{-}

Check the count using vanadium’s atomic number and the 18-electron argon core: $N_{electrons} = 23 - 2 = 21$ $N_{3 d} = 21 - 18 = 3$

By Hund’s rule, these three electrons occupy separate d orbitals before pairing: $N_{unpaired} = 3$

Which has the higher third ionization enthalpy, chromium or manganese?

Manganese has the higher third ionization enthalpy. Now the starting species are the doubly charged ions:

C r^{2 +} : [A r] 3 d^{4} \to C r^{3 +} : [A r] 3 d^{3} + e^{-}

M n^{2 +} : [A r] 3 d^{5} \to M n^{3 +} : [A r] 3 d^{4} + e^{-}

Manganese now loses an electron from a half-filled subshell, making its third ionization enthalpy higher. Changing the ionization number changes the starting species: write the starting-ion configurations again rather than carrying over the second-ionization order.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

For a worked example of the same idea, see Gravitation NEET 2011: When Is Gravitational Power Greatest?.

Frequently asked questions

What is the second ionization enthalpy order of Ti, V, Cr and Mn?

The decreasing order is Cr > Mn > V > Ti, so option A is correct. Chromium ranks highest because its second ionization removes a 3d electron from the stable half-filled 3d5 configuration of Cr+.

Why is chromium's second ionization enthalpy higher than manganese's?

Cr+ has the configuration [Ar] 3d5, so removing its next electron breaks a half-filled 3d subshell. Mn+ has [Ar] 3d5 4s1 and loses its remaining 4s electron, retaining the half-filled 3d5 subshell.

Which ion is the starting species for second ionization enthalpy?

Second ionization starts from the singly charged gaseous ion, M+(g), and produces M2+(g) and an electron. It does not mean removing two electrons together from a neutral atom.

Why is manganese above vanadium and titanium in second ionization enthalpy?

Mn+, V+ and Ti+ each lose a remaining 4s electron during second ionization. For these species, the nuclear-charge comparison places manganese above vanadium and vanadium above titanium. Forming a stable half-filled Mn2+ ion does not make manganese the lowest in the set.

Which has higher third ionization enthalpy, chromium or manganese?

Manganese has the higher third ionization enthalpy. Mn2+ starts with [Ar] 3d5, so its third ionization breaks a half-filled subshell, whereas Cr2+ starts with [Ar] 3d4. The second-ionization ranking cannot simply be carried over.