Q.What is ionization enthalpy?

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 $\approx 906$.

Zinc has the electronic configuration $[Ar]3d^{10}4s^2$ — both the 3d and 4s subshells are completely filled, giving it extra stability. Removing an electron from this fully-filled, symmetrical arrangement requires more energy than removing an electron from Sc, Co or Cu, whose d-subshells are not completely filled. Hence Zn has the highest first ionization enthalpy of the four.

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).
• Chalcogens (Group 16) and alkaline earth metals (Group 2) have intermediate/higher ionisation enthalpies than alkali metals, since alkaline earth metals have a filled/stable ns2 configuration that is somewhat harder to disturb than the lone ns1 electron of alkali metals, and are also to the right of alkali metals in the same period.

So alkali metals consistently show the lowest ionisation enthalpy values in the periodic table.

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 2s2, despite B having one more proton.

So although the general trend is Li < Be < B < C, the extra stability of Be's filled subshell makes IE(Be) > IE(B). This gives the true experimental order: Li (≈520) < B (≈801) < Be (≈899) < C (≈1086) kJ/mol, i.e., decreasing order C > Be > B > Li.

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 of Be's 2s electrons — giving B a lower first ionisation energy than Be, despite the general left-to-right increasing trend.

So IP(Be) = 9.32 eV, IP(B) = 8.29 eV.

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
• ns2np6 (noble gas configuration): p subshell COMPLETELY filled — the most stable possible arrangement, requiring the most energy of all to disturb by removing an electron

Between the half-filled (np3) and fully-filled (np6) special stabilities, the fully-filled noble-gas configuration is more stable overall (this is why noble gases have the highest ionisation energies of any elements in their respective periods).

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 atoms in their period to ionize, i.e. lowest IE1. (They do NOT have the smallest radius, highest IE1, or highest electronegativity — those describe elements at the other end of the period, near the halogens/noble gases.)

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 — ionisation energy goes up. This is exactly why ionisation energy generally increases from left to right across a period (nuclear charge increases while shielding stays roughly constant).

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.
• ns2 np3 is a half-filled p-subshell (extra stable due to symmetric distribution and exchange energy), giving a local maximum, but still lower than a full octet.
• ns2 np1 and ns2 np2 are partially filled, less stable, and have comparatively lower ionization energies.

Among the four, the complete octet ns2 np6 has the 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. Hence IE2 is much greater than IE1 for alkali metals.