Terbium (Z = 65) is a lanthanoid and all others are actinoids. Lanthanoids have atomic numbers from 58 to 71. Actinoids have atomic numbers from 90 to 103.

4f and 5f orbitals are not equally shielded. 5f are poorly shielded as 5f orbitals are away from the nucleus.

LANTHANIDES

LANTHANIDES

Lu, Gd and La belongs to lanthanide series and Th belongs to actinide series.

We have 15 elements in actinides. They are from atomic number 89 to 103 Actinium, Thorium, Protactinium, Uranium, Neptunium, Plutonium, Americium, Curium, Berkelium, Californium, Einsteinium, Fermium, Mendelevium, Nobelium, Lawrencium. Hence, from these list we could say that Actinium is the first and Lawrencium is the last element in actinides.

Actinides are also called the 5f series.

$5f$ orbitals are bigger than $4f$ orbitals, in other words, $5f$ orbitals are more diffused thereby having the poor shielding effect. Hence have a more effective nuclear charge thereby having a small size. Hence actinide contraction is more than lanthanide contraction.

The two rows that are generally placed underneath the main Periodic Table are called the lanthanide series and the actinide series.

These two rows are produced when electrons are being added to f orbitals.

Therefore, this block of elements are referred to as the "f block".

Actinide series starts with actinium (Z=89) and contains fourteen elements from thorium (Z=90) to lawrencium (Z=103). This is followed by Rutherfordium (Z=104). There Rutherfordium (Z=104) also known as Unnilquadium.

Am shows +2, +3, +4, +5, +6, +7 oxidation states.

Oxidations are the polyatomic cations contain one or more oxygen atoms. Actinides have a strong tendency towards the complex formation and form oxocations like oxides and hydroxides and are basic in nature.

The actinide contraction is due to imperfect shielding of $5f$ electron. The valence shell electronic configuration of actinoids is $\displaystyle 5f^{1-14}6d^{0-1}7s^2$. With an increase in atomic number, the positive charge on the nucleus increases by $+1$, and one more electron enters in the same $5f$ subshell. There is imperfect shielding of one electron by another electron in the same $5f$ subshell. Due to this, the atomic and ionic radii show a gradual decrease with increase in atomic number.

The name of californium was derived from the name of the state and University of California.

Beryllium resembles Aluminium in properties. This is due to similar electronegativity, similar atomic size and they diagonal relationship.

Amphoteric oxides can behave as an acid as well as base. They react with both acids and bases to form salts. Tin oxide although doesn't dissolve in water but it is amphoteric in nature. Zinc oxide is also amphoteric in nature.

E.A of $Li^{+3} = -$I.P of $Li^{+2}$
I.P of $Li^{+2}  = $I.P of $H \dfrac{Z^2}{n^2}$
Therefore, I.P of $Li^{+2}$=$2.18 \times 10^{-18} J/atom \times 9 $ as Z=3
Thus, E.A of $Li^{+3} =-1.962 \times 10^{-17}$ J/atom

As the seventh period will be completely filled its electronic configuration must be in the manner of noble gas configuration i.e., $ns^2,: np^6$ because the last element in every period is noble gas element. Hence, the electronic configuration of last element of seventh period is (Rn) $5f^{14} 6d^{10} 7s^27p^6$ as the electronic configuration contains $118$ electrons, the last element in the seventh period will have the atomic number $118$.

Generally the elements having  half filled and completely filled outer electronic configurations are stable  elements,here in given options the configuration $1s^22s^22p^63s^23p^64s^23d^5$ contains half filled outer electronic configuration $(3d^5)$.

$Z=106$ is seaborgium and its electronic configaration is $(Rn) 5f^{14}6d^{4}7s^{2}$. as the differenting electron enter into $7s$, 7 is the principle quantum number which represents the number of period to which the element belongs hence, the element belongs to 7th period.

The total number of electrons present in all the shells are $2+8+3=13$, hence the atomic number of element is $13$ and the electronic configuration would be $1s^22s^22p^63s^23p^1 $. Element belongs to $|||A$ group.