Chemistry Test - 4

Sucrose is a molecule composed of two monosaccharides, namely glucose and fructose. This non-reducing disaccharide has a chemical formula of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\).

In a \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) molecule, the fructose and glucose molecules are connected via a glycosidic bond. This type of linking of two monosaccharides called glycosidic linkage. Sucrose has a monoclinic crystal structure and is quite soluble in water. It is characterized by its sweet taste.

The beryllium fluoride cannot be easily prepared by simple methods. It is known that beryllium oxide dissolves in aqueous hydrofluoric acid with the formation of the fluoride, but on evaporation of the resulting solution some of the combined acid is lost, and the residue is believed to be an oxy fluoride.

Thermal decomposition of \(\left(\mathrm{NH}_4\right)_2 \mathrm{BeF}_4\) is the best route for the preparation of \(\mathrm{BeF}_2\)

\(\left(\mathrm{NH}_4\right)_2 \mathrm{BeF}_4 \rightarrow 2 \mathrm{NH}_4 \mathrm{~F}+\mathrm{BeF}_2\)

The \(\Delta \mathrm{H}_{\mathrm{f}}\) of alkali metal halides is shown through this graph.

This graph shows that formation of metal halide is negative, i.e. energy is released during the formation of metal halides.

(A)

\(\mathrm{M}-\mathrm{F} \)

\(\mathrm{M}-\mathrm{Cl} \)

\(\mathrm{M}-\mathrm{Br} \)

\(\mathrm{M}-\mathrm{I}\)

On moving down the group, the size of halogens increase which results in the lengthening of bond, making the bond weaker. Therefore, \(\Delta \mathrm{H}_{\mathrm{f}}\) is less negative on moving down the group as the stability of metal halide decreases down the group.

\(\text { (B) } \)

\(\mathrm{Li}-\mathrm{X} \)

\(\mathrm{Na}-\mathrm{X} \)

\(\mathrm{K}-\mathrm{X} \)

\( \mathrm{Rb}-\mathrm{X}\)

On moving down the group, the electropositivity of metal increases making the attraction between the oppositely charged ions even stronger, thus increasing the strength of bond. Increase in strength is more than the weakness produced due to lengthening of bond because of increase in size of metal ions down the group. This makes \(\Delta \mathrm{H}_{\mathrm{f}}\) more negative on moving down the group. Option (a) is incorrect, due to reason produced in (B). Option (b) is incorrect. As CsI has \(\mathrm{Cs}^{\oplus}\) and \(\mathrm{I}^{\ominus}\) an constituent ions and both the ions are larger in size. The lattice enthalpy is not so high as shown in graph. But it has low solubility due to less hydration energy released when hydration of larger ions takes place. Option (c) is correct. As LiF has highest lattice enthalpy as shown in graph. This is due to the size of ions that is smallest in their respective groups. Smaller is the size, shorter is the bond length and stronger is the bond. The energy required is break the bond and release the constituents ions is more than the energy released in the hydration of ions. So, LiF is least soluble in water. Option (d) is incorrect. LiF has most negative enthalpy in all metal fluorides as shown in graph.

Given reaction is:

\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{Fe}^{2+}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{Cr}^{3}+\mathrm{Fe}^{3+}+\mathrm{CO}_{2}\)

The reaction in balanced form will be as follows:

\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+2 \mathrm{Fe}^{2+}+2 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{Cr}^{3+}+2 \mathrm{Fe}^{3+}+4 \mathrm{CO}_{2}\)

The oxidation number of chromium in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is \(+6\) and it reduces to \(+3\) in \(\mathrm{Cr}^{3+}\). On balancing the equation, we will see 2 moles of chromium ion goes from \(+6\) to \(+3\).

Therefore, there are 6 electrons involved in the above redox reaction.

More the number of upaired \(\mathrm{d}\)-electrons, more is the magnetic moment.

Therefore, we have:

(A) \(\mathrm{Cu}^{2+}: 3 \mathrm{d}^{9}\)

No. of unpaired electrons \(=1\)

(B) \(\mathrm{Ni}^{2+}: 3 \mathrm{d}^{8}\)

No. of unpaired electrons \(=2\)

(C) \(\mathrm{Co}^{2+}: 3 \mathrm{d}^{7}\)

No. of unpaired electrons \(=3\)

(D) \(\mathrm{Fe}^{2+}: 3 \mathrm{d}^{6}\)

No. of unpaired electrons \(=4\)

Therefore, \(\mathrm{F e}^{2+}\) has highest magnetic moment.

A solution of acetone in ethanol shows a positive deviation from Raoult's law.

• It is due to miscibility of these two liquids with a difference of polarity and length of the hydrocarbon chain. 
• Positive derivation occurs when vapour pressure of the component is greater than expected value.
• Acetone and ethanol both the components escape easily showing higher vapour pressure than the expected value.

As we move from top to bottom in a group, the number of valence electrons remains same because a group is defined such that their valence shell configuration is same. So, valence electrons remain same but valence shell changes.

We shall convert all values into grams

\(40 \mathrm{~g}\) of \(\mathrm{Fe}=40 \mathrm{~g}\)

6 moles of \(\mathrm{N}_{2}\) at NTP

We know that,

Moles \(=\frac{\text { weight in grams }}{\text { molecular Weight }}\)

Weight of \(\mathrm{N}_{2}=6 \times\) Molecular mass of \(\mathrm{N}_{2}\)

Weight of \(\mathrm{N}_{2}=6 \times 28=168 \mathrm{~g}\)

\(0.2 \mathrm{~g}\) of silver \(=0.2 \mathrm{~g}\)

\(10^{23}\) atoms of carbon

1 mole of carbon \(=6.022 \times 10^{23}\) atoms \(=12 \mathrm{~g}\)

\(10^{23}\) atoms \(=\frac{10^{23}}{6.023 \times 10^{23}} \times 12=1.992 \mathrm{~g}\)

Thus, the highest mass is of 6 moles of \(\mathrm{N}_{2}\) at NTP.

Neutral ligand means ligand with no charge on it.

Example: H₂O, NH₃, CO, C₂ H₄...

ONO^- has a charge on it, therefore it is not a neutral ligand.

Given compound:

\(\mathrm{\underset { \overset { | }{ CHO }  }{ { CH }_{ 2 } } -\underset { \overset { | }{ CHO }  }{ CH } -\underset { \overset { | }{ CHO }  }{ { CH }_{ 2 } }}\)

Numbering the carbon atoms, we get:

\(\mathrm{\underset { \overset { | }{ CHO }  }{ { \overset {3}{C}H }_{ 2 } } -\underset { \overset { | }{ CHO }  }{\overset {2}{C}H } -\underset { \overset { | }{ CHO }  }{ { \overset {1}{C}H }_{ 2 } }}\)

• Since there are three carbon atoms in long chain and it contains 3 -CHO groups (one on each carbon atom). 
• IUPAC suffix "aldehyde" will added due to presence of functional group -CHO.

Therefore, the correct IUPAC name will be propane-1, 2, 3-tricarbaldehyde.