Q.Which of the following will have highest fourth ionization enthalpy IE₄? La⁴⁺, Gd⁴⁺, Lu⁴⁺.

Zn has the highest first ionization enthalpy among Zn, Cu, Co, Sc because of its extra-stable fully-filled $3d^{10}4s^2$ configuration.

Approximate first ionization enthalpies (kJ/mol): Sc $\approx 633$, Co $\approx 760$, Cu $\approx 745$, Zn $\approx 906$.

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Ionisation enthalpy generally decreases down a group and increases across a period. Alkali metals (Group 1) sit at the far left of the table with a single, loosely-held valence electron, giving them the lowest ionisation enthalpies overall.

Ionisation enthalpy is the energy needed to remove the outermost electron from a gaseous atom. Alkali metals (Li, Na, K, Rb, Cs) have just one electron in their outermost shell, which is well-shielded from the nucleus by inner shells and is easily removed -- this is why they are so reactive and readily form +1 cations.

Compared to the other options:

• Halogens (Group 17) have HIGH ionisation enthalpy (they tend to gain electrons, not lose them). …

The correct decreasing order of first ionization enthalpy is C > Be > B > Li, because Be's filled 2s2 configuration makes it anomalously harder to ionize than B.

Across Period 2, ionization enthalpy generally increases left to right as nuclear charge increases and atomic radius decreases (Li → Be → B → C → ...). However there is a well-known dip at boron: Be has the electron configuration [He]2s2 — a completely filled 2s subshell, which is a stable, symmetrical arrangement. Removing an electron from this filled subshell requires extra energy. B has configuration [He]2s2 2p1 — its single 2p electron is at slightly higher energy and is more shielded by the inner 2s2 pair (poor penetration of p-orbital compared to s), so it is actually EASIER to remove than an electron from Be's filled 2s …

Be's first IP (9.32 eV) is higher than B's (8.29 eV), even though B comes after Be in the periodic table.

Across Period 2, ionisation energy generally increases left to right, but Be (2s2, fully filled subshell) is exceptionally stable, so removing an electron from it requires more energy. Boron's outermost electron sits in a 2p orbital, which is higher in energy and better shielded by the 2s2 electrons, so it is removed more easily than one …

Fully-filled (ns2np6) and half-filled (ns2np3) configurations both have extra stability, but a completely filled octet is more stable than a half-filled one, so ns2np6 has the higher ionisation energy of the two 'special' cases — and both are higher than the partially-filled np1/np2 cases.

Ionisation energy generally increases with the stability of the resulting configuration after electron removal being LESS favourable (i.e. the more stable the starting configuration, the harder it is to pull an electron out of it).

Compare the four given outer configurations:

• ns2np1 (e.g. boron family): only 1 electron in p subshell, relatively easy to remove -> lower IE
• ns2np2 (e.g. carbon family): 2 electrons in p subshell, no special extra stability -> moderate IE
• ns2np3 (e.g. nitrogen family): p subshell exactly HALF-filled (one electron in each of the 3 p orbitals, per Hund's rule) — this symmetric arrangement has extra stability, giving nitrogen an anomalously high IE compared to its neighbour oxygen …

Alkali metals sit at the far left of each period, so they have the largest atomic size and least tightly held valence electron among that period's elements — hence the lowest first ionization energy (IE1).

Ionization energy generally increases left-to-right across a period as effective nuclear charge rises and atomic radius shrinks (electrons held more tightly). Alkali metals (Li, Na, K, Rb, Cs — Group 1) are the first element of each period, with only one loosely-held electron in an outer s-orbital, shielded by the underlying noble-gas core. This makes them the easiest …

Ionisation energy increases with increasing nuclear charge, because a stronger positive charge holds the outer electrons more tightly.

Ionisation energy is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom. When the nuclear charge (number of protons) increases while the number of shells/screening stays roughly the same, the effective nuclear charge experienced by the outermost electrons increases. This pulls the valence electrons closer and binds them more strongly to the nucleus, so more energy is required to remove an electron — ionis …

Fully filled and half-filled configurations resist electron removal; ns2 np6 is a complete noble-gas shell, the most stable of the four options.

Ionization energy trends with the stability of the electron configuration:

• ns2 np6 is a fully filled p-subshell (noble-gas-like, e.g. neon/argon configuration) -- extra stable, hardest to remove an electron from -- highest ionization energy. …

The second ionization energy of an alkali metal is much higher than the first because, after losing one electron, the metal ion attains a stable noble-gas electron configuration.

An alkali metal atom (e.g., Na: [Ne] 3s1) has a single electron in its outermost s-orbital, which is well shielded and loosely held, so it is removed relatively easily (low IE1). Once this electron is removed, the resulting M+ ion (e.g., Na+: [Ne]) has a completely filled, stable noble-gas configuration. Removing a second electron means breaking into this stable, filled shell, which is both smaller in size (higher effective nuclear charge per electron) and electronically stable — this requires a much larger amount of energy …