Chemistry Test - 7

J J Thomson, in 1898, proposed the plum pudding, (raisin pudding or watermelon) model of an atom. An important feature of this model is that the mass of the atom is assumed to be uniformly distributed over the atom. This model was able to explain the overall neutrality of the atom.

According to Thomson model of an atom:

• An atom has a spherical shape with radius \(\approx 10^{-10} m\).
• The positive charge is uniformly distributed in the atom.
• The electrons are embedded such that atom posses the most stable electrostatic arrangement. Thereby the model was able to explain the overall neutrality of the atom.
• The mass of the atom is assumed to be uniformly distributed over the atom. It could not explain the atomic spectra, stability of the atom and the actual distribution of electrons, protons and neutrons in the atom.

Each water molecule can form a maximum of four hydrogen bonds with neighboring water molecules. The two hydrogens of the water molecule can form hydrogen bonds with other oxygens in ice, and the two lone pairs of electrons on the oxygen of the water molecule can attract other hydrogens in ice. So, 4 hydrogen bonds are possible witha molecule of water.

Here, \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) with dissociation constant \(1.9 \times 10^{17}\) is more stable than \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) with dissociation constant \(2.6 \times 10^{37}\).

Higher is the dissociation constant lesser will be the stability of complex. So\(\mathrm{B}\) is more stable than \(\mathrm{A}\).

Given reaction,

\(4 \mathrm{Fe}+3 \mathrm{O}_{2} \longrightarrow 4 \mathrm{Fe}^{+3}+6 \mathrm{O}^{2-}\)

Clearly, given reaction is a redox reaction in which, Fe gets oxidised to \(\mathrm{Fe}^{+3}\) by losing of three electrons, and \(O_{2}\), gets reduced to \(O^{2-}\).

Thus, Metalliciron \((F e)\) is a reducing agent and \(\mathrm{Fe}^{+3}\) is an oxidising agent.

Volume of \(0.5\) mole of steam at \(1 atm\) pressure

\(=\frac{ nRT }{ P }=\frac{0.5 \times 0.0821 \times 373}{1.0}=15.3 L\)

Change in volume \(=\) vol. of steam \(-\) vol. of water \(=15.3-\)negligible \(=15.3 L\)

Work done by the system,

\( = P _{\text {ext }} \times \text { volume change }\)

\(=1 \times 15.3=15.3 \text { litre-atm } \)

\( =15.3 \times 101.3 J =1549.89 J\)

'W' should be negative as the work has been done by the system on the surroundings.

W \(\mathrm{=-1549.89 J =-1.54 kJ}\)

The maximum covalent character is shown by the compound is AlCl₃.

The proportion of covalent character in an ionic bond is decided by polarisability of the metal cation as well as the electronegativity of both elements involved in bonding. Polarisability is further decided by the density of positive charge on the metal cation. AICI₃ is considered to show maximum covalent character among the given compounds. This is because Al³⁺ bears 3 unit of positive charge and shows strong tendency to distort the electron cloud, thus the covalent character in Al-CI bond dramatically increases.

By the below given relation:

\(\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { Salt }]}{[\text { Base }]}\)

when \([\) Salt \(]=[\) Base \(]\)

And \(\log 1 = 0\)

So, \(\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}=4.75\)

So, \(\mathrm{pH}=14-4.75=9.25 \)

Nickel (Ni) element forms interstitial compounds.

• It is known that a compound which is formed from a small atom whose radius is small enough that it sits in an interstitial "hole" in a metal lattice is known as an interstitial.
• When we move from left to right in a period then there occurs a decrease in size of the atoms. And, atoms with smaller size are able to sit in an interstitial hole.
• Out of the given options nickel is smaller in size. As a result it will be able to form an interstitial compound, whereas, Sc, Fe and Co are all larger in size than nickel. Therefore, they will not form interstitial compounds.

Among the following mixtures, dipole-dipole as the major interaction is present inAcetonitrile and acetone.

\({MgCl}_{2} \cdot 6 {H}_{2} {O}\) and \({CaCl}_{2} \cdot 6 {H}_{2} {O}\)

Onheating hydrated \({CaCl}_{2}\) is dehydrated while hydrated \({MgCl}_{2}\)changes into \({MgO}\)

Onheating hydrated \({CaCl}_{2}\) is dehydrated while hydrated \({MgCl}_{2}\)changes into \({MgO}\)

\({CaCl}_{2}\cdot 6 {H}_{2} {O} \stackrel{\text { heat }}{\longrightarrow} {CaCl}_{2}+6{H}_{2} {O}\)

\({MgCl}_{2}\cdot 6 {H}_{2} {O} \stackrel{\text { heat }}{\longrightarrow} {MgO}+2 {HCl}+5{H}_{2} {O}\)