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Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

Most of the radio active elements are present in__________

  1. Lanthanoids

  2. Actinoids

  3. Representative elements

  4. Second transition series
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The actinoids are radioactive elements and the earlier members have relatively long half-lives.

The most common and known element is Uranium, which is used as nuclear fuel when its converted into plutonium, through a nuclear reaction.

Option B is correct.
Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

The actinide contraction is due to:

  1. perfect shielding of $5f$ election
  2. imperfect shielding of $4f$ election
  3. imperfect shielding of $5f$ electron
  4. perfect shielding of $4f$ election
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

Which of the following is true regarding derivation of the name of californium?

  1. The name of californium was derived from the name of the place.

  2. The name of californium was derived from its color.

  3. The name of californium was derived from the name of the scientist who discovered it.

  4. The name of californium was derived from mythological character name.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

Beryllium resembles Aluminium in properties. This is mainly due to:

  1. equal electro negativity values of elements

  2. equal atomic volumes of the elements

  3. equal electron affinity

  4. equal nuclear charges in their atoms
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

Amphoteric behaviour is shown by the oxides of :

  1. $Al$ and $Ca$
  2. $Pb$ and $Ba$
  3. $Cr$ and $Mg$
  4. $Sn$ and $Zn$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

$ZnO+HCl \rightarrow ZnCl _2+H _2O$
$ZnO+NaOH \rightarrow Na _2ZnO _2+H _2O$
$SnO _2+HCl \rightarrow SnCl _4+H _2O$
$SnO _2+NaOH+H _2O \rightarrow Na _2[Sn(OH) _6]$

Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

If the IP of hydrogen in its ground state is 2.18 x$10^{-18}$ J/atom, then the electron affinity of $Li^{3+}$ ion is :

  1. $-2.18\times$$10^{-18}$J/atom
  2. $-6.54 \times$$10^{-18}$J/atom
  3. $-3.488 \times$$10^{-18}$J/atom
  4. $-1.962\times$$10^{-17}$ J/atom
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

If the seventh period is completed, the atomic number of the last element would be:

  1. $118$
  2. $112$
  3. $107$
  4. $120$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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 $118$.

Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

Stable electronic configuration of a transition metal is:

  1. $1s^{2} 2s^{2}2p^{6} 3s^{2} 3p^{6} 3d^{9} 4s^{2}$
  2. $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{5}$
  3. $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2}$
  4. $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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 $(3d^5)$.


Hence option B is correct.

Multiple choice chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

The element with $Z = 106$ belongs to :

  1. $3^{rd}$ period
  2. $ 5^{th}$ period
  3. $7^{th}$ period
  4. $6^{th}$ period
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Z=106$ 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.