Inorganic Chemistry Questions and Answers

Answer: B

The size order is \(\text{F}^{-} \gt \text{Na}^{+} \gt \text{Mg}^{2+} \gt \text{Al}^{3+}\).

All four ions are isoelectronic, with 10 electrons each (the neon configuration). For isoelectronic ions, size falls as nuclear charge rises, because the same electron cloud is pulled in more strongly. The nuclear charges are F 9, Na 11, Mg 12, Al 13, so F^- is largest and Al³⁺ smallest.

The orders starting with Al³⁺ reverse the trend, and the order putting Na⁺ before F^- ignores that fluoride has two fewer protons for the same ten electrons.

Answer: B

The least second ionization energy is that of Si.

The second ionization energy removes an electron from the singly charged ion, so compare the configurations of the M⁺ ions: Al⁺ is [Ne]3s², Si⁺ is [Ne]3s²3p¹, P⁺ is [Ne]3s²3p² and S⁺ is [Ne]3s²3p³.

Si⁺ loses a lone 3p electron, which is shielded by the filled 3s pair and is easy to remove. Al⁺ must lose an electron from the filled 3s subshell, which is more penetrating and more tightly held, so aluminium's second ionization energy is higher than silicon's even though aluminium has the lower first ionization energy.

P⁺ and S⁺ have greater nuclear charge, and S⁺ also has the stable half-filled 3p³ set, so both need more energy than Si⁺.

Answer: D

The correct order is S \(\gt\) P \(\gt\) Si \(\gt\) Al.

Across a period, electronegativity rises with nuclear charge, because the atoms shrink and hold their valence electrons more tightly. Al, Si, P and S are consecutive third-period elements, so electronegativity increases in that order; the Pauling values are roughly Al 1.6, Si 1.9, P 2.2, S 2.6.

Al \(\gt\) Si \(\gt\) P \(\gt\) S is the exact reverse. The two mixed orders put Si above P or P below Al, which no scale supports.

Answer: D

The correct statement is Ionic radius order is K⁺ > Ca²⁺ > Mg²⁺.

K⁺ and Ca²⁺ are isoelectronic (18 electrons); Ca²⁺ has one more proton, so it pulls the same electrons in more tightly and is smaller. Mg²⁺ has one shell fewer and is smaller still.

The other statements are wrong. Electron affinity runs Cl > F > O: fluorine's small, crowded 2p shell makes its electron affinity lower than chlorine's. First ionisation energy runs Mg > Al > K, because Al loses an easily removed 3p electron while Mg must lose one from a filled 3s subshell. Atomic radius increases down the group, so As > P > N.

Answer: B

The correct decreasing order is Cl > Br > I > At.

Electron gain enthalpy becomes less negative down group 17 because the added electron enters a shell further from the nucleus and is held less firmly. Chlorine releases about 349 kJ/mol on gaining an electron, bromine about 325, iodine about 295, and astatine less still.

Chlorine heads the group, rather than fluorine, because fluorine's small 2p shell is so crowded that electron repulsion reduces its value; fluorine is not in this list, so chlorine is simply first. The options that put bromine, iodine or astatine ahead of chlorine reverse the down-the-group trend.

Answer: A

The metalloid is antimony.

Metalloids lie along the boundary between metals and non-metals in the p-block and show intermediate properties. The usual list is boron, silicon, germanium, arsenic, antimony and tellurium.

Aluminium, magnesium and calcium are metals, and phosphorus is a non-metal.

Answer: C

Main group elements are both s- and p-block elements.

The representative or main group elements are those of groups 1, 2 and 13 to 18, in which the last electron enters an s or p orbital of the outermost shell.

The d-block elements are the transition elements and the 4f and 5f elements are the inner transition elements (lanthanoids and actinoids); neither is a main group. Options naming only the s-block or only the p-block are incomplete.

Answer: B

The largest positive electron gain enthalpy belongs to Ne (neon).

Noble gases have completely filled valence shells, so an added electron must enter the next, higher shell. Energy has to be supplied, which is why their electron gain enthalpies are positive.

The standard table gives approximately He +48, Ne +116, Ar +96, Kr +96, Xe +77 and Rn +68 kJ mol^-1. Neon is highest: its incoming electron has to go into the 3s orbital of a small atom whose 2p shell is full, which is strongly unfavourable. Helium is smaller still, but its value is much lower, and from argon down the larger atoms accept the extra electron more easily, so Ar, Kr and Rn all fall below neon.

Answer: E

Element 105 is Dubnium (Db).

The heavy actinoids and the first transactinoids run in order: Mendelevium (101), Nobelium (102), Lawrencium (103), Rutherfordium (104) and Dubnium (105). IUPAC's temporary systematic name for 105 was unnilpentium (Unp), from the digits 1-0-5, before the permanent name dubnium was adopted after the Russian research centre at Dubna.

The other four options are the elements immediately before it, a common trap when the numbers are remembered only roughly.

Answer: A

The correct order is Na \(\gt\) Mg \(\gt\) Si \(\gt\) P.

Across a period from left to right, the nuclear charge rises and atoms hold their electrons more tightly, so the tendency to lose electrons, metallic character, falls. These four are in period 3 in the order Na, Mg, Si, P, so metallic character decreases in that order.

Putting Mg before Na, or P before Si, breaks the periodic trend, and the last two options reverse it.