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Multiple choice chemistry what is inside atom discovery of neutrons structure of atoms discovery of subatomic particles

Who determined the mass of a neutron?

  1. John Dalton (early 1800's)

  2. Ernest Rutherford (1911)

  3. Niels Bohr (1913)

  4. James Chadwick (1932)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The presence of neutron and its mass was determined by James Chadwick. Neutron was discovered by bombarding $\alpha-$particles on $Be$. On doing so, highly energetic beam of radiation was observed by Chadwick. Thus mass of neutron was determined by James Chadwick.

Multiple choice chemistry what is inside atom discovery of neutrons structure of atoms discovery of subatomic particles

Whose experiment proved the existence of the nucleus?

  1. John Dalton (early 1800's)

  2. Ernest Rutherford (1911)

  3. Niels Bohr (1913)

  4. James Chadwick (1932)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The $\alpha-$ ray scattering-gold foil experiment by Ernest Rutherford gave substantial evidence for  existence of nucleus in $1911$. Thus the existence of nucleus was proved by Rutherford's experiment.

Multiple choice chemistry what is inside atom discovery of neutrons structure of atoms discovery of subatomic particles

Neutron was discovered by:

  1. J.J. Thomson

  2. Chadwick

  3. Rutherford

  4. Priestley
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Neutron was discovered by Chadwick.
To overcome the failure of proton electron hypothesis, Rutherford, in 1920, proposed the existence of neutral particle made from close combination of proton and electron. He named it neutron.
In 1932, Lord Chadwick proved the existence of neutron by bombarding lighter elements (boron and beryllium) with alpha particles which resulted in the emission of highly penetrating neutral radiation.

Multiple choice chemistry what is inside atom discovery of neutrons structure of atoms discovery of subatomic particles

The neutron is a neutral (non-electrical) particle in the centre of the atom. It was discovered when a tiny piece of the metal, beryllium, was bombarded with alpha particles. This discovery was made by:

  1. Chadwick

  2. Rutherford

  3. Niels Bohr

  4. Enrico Fermi
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
In 1930, it was discovered that Beryllium, when bombarded by alpha particles, emitted a very energetic stream of neutral radiation. In 1932, Chadwick proposed that neutral radiation consists of a new type of neutral particles - neutrons.
Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

A translucent white waxy solid (A) reacts with excess of chlorine to give a yellowish white powder (B). (B) reacts with organic compounds containing -OH group converting them into chloro derivatives. (B) on hydrolysis gives (C) and is finally converted to phosphoric acid. (A), (B) and (C) are?

  1. $P _{4}$,$PCl _{3}$,$H _{3}PO _{4}$
  2. $P _{4}$,$PCl _{5}$,$H _{3}PO _{3}$
  3. $P _{4}$,$PCl _{5}$,$POCl _{3}$
  4. $P _{4}$,$PCl _{3}$,$POCl _{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A translucent white waxy solid (A) reacts with excess of chlorine to give a yellowish white powder (B). Here B is $PCl _5$.$PCl _5$ reacts with organic compounds containing -OH group converting them into chloro derivative $POCl _3. $.$PCl _5$ on hydrolysis gives (C).C is $POCl _3$ and is finally $POCl _3$ converted to phosphoric acid.

So  (A), (B) and (C) are $P _{4}$,$PCl _{5}$,$POCl _{3}$.

Hence option C is correct.

Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

$\mathrm{P}\mathrm{Cl} _{3}$ is prepared by the action of $\mathrm{Cl} _{2}$ on :

  1. $ \mathrm{P} _{2}\mathrm{O} _{3}$
  2. $ \mathrm{P} _{2}\mathrm{O} _{5}$
  3. White $\mathrm{P}$
  4. $\mathrm{H} _{3}\mathrm{P}\mathrm{O} _{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

White P is reactive due to strain.
$P _{4}+6Cl _{2}\ \to \ 4PCl _{3}$
(White P)

Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

What is the oxidation state of Phosphorus in ${P}{Cl} _{3}$?

  1. ${+}{5}$
  2. ${+}{3}$
  3. ${+}{1}$
  4. ${-}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The oxidation state of ${P}$ in ${P}{Cl} _{3}$ is ${+3}$.
Each Chlorine atom has an oxidation number of ${-1}$. There are three chlorine atoms, hence the total oxidation number of the Chloride ions in ${-3}$. ${PCl} _{3}$ molecule is neutral. Hence the oxidation number of Phosphorus should be ${+3}$

Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

What is the hybridization of Phosphorus Trichloride?

  1. ${sp}^{3}$
  2. ${sp}^{2}$
  3. ${sp}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The hybridization of ${PCl} _{3}$ is ${sp}^{3}$. Phosphorus has five electrons in its outermost shell. Thus it forms three sigma bonds with three neighbouring Chlorine atoms. The remaining two electrons remain as a lone pair on Phosphorus atom. Thus these three sigma bonds and one lone pair, make this molecule ${sp}^{3}$ hybridized.