Which of the following is incorrect statement?

Question

$\textbf{Reduction potential of ions are given below:}$
\[ \begin{array}{ccc} \text{ClO}_4^- & \text{IO}_4^- & \text{BrO}_4^- \\ E^\circ = 1.19 \, \text{V} & E^\circ = 1.65 \, \text{V} & E^\circ = 1.74 \, \text{V} \\ \end{array} \]
The correct order of their oxidising power is:

Answer 1

To determine the incorrect statement, we need to evaluate the nature of each compound provided in the options. Let's go through each one:

$PbF_4$ is covalent in nature
- Lead fluoride, $PbF_4$, is generally considered ionic in nature due to the significant difference in electronegativity between lead (Pb) and fluorine (F). However, due to the presence of a highly electronegative fluorine and the higher oxidation state of lead (+4), some degree of covalency can be present. Nonetheless, it’s commonly considered to have ionic characteristics.
- Thus, the statement claiming it is covalent without qualification is incorrect.
$SiCl_4$ is easily hydrolysed
- Silicon tetrachloride, $SiCl_4$, is indeed easily hydrolyzed by water.
- This is due to silicon’s ability to form strong bonds with oxygen. When $SiCl_4$ comes into contact with water, it reacts to form silicic acid and hydrochloric acid.
$GeX_4$ (X = F, Cl, Br, I) is more stable than $GeX_2$
- Germanium tetrahalides $GeX_4$ are more stable than germanium dihalides $GeX_2$.
- This increased stability is due to germanium being in a higher oxidation state (+4) compared to the lower oxidation state (+2) in $GeX_2$, thereby providing greater stability.
$SnF_4$ is ionic in nature
- Tin(IV) fluoride, $SnF_4$, has a high degree of ionic character due to the presence of highly electronegative fluorine atoms and the +4 oxidation state of tin, leading to a weakened bond covalency.
- Thus, this statement is correct as $SnF_4$ is largely ionic in nature.

After evaluating all options, the statement that is incorrect is: $PbF_4$ is covalent in nature.

Answer 2

To determine the incorrect statement, we need to evaluate the nature of each compound provided in the options. Let's go through each one:

$PbF_4$ is covalent in nature
- Lead fluoride, $PbF_4$, is generally considered ionic in nature due to the significant difference in electronegativity between lead (Pb) and fluorine (F). However, due to the presence of a highly electronegative fluorine and the higher oxidation state of lead (+4), some degree of covalency can be present. Nonetheless, it’s commonly considered to have ionic characteristics.
- Thus, the statement claiming it is covalent without qualification is incorrect.
$SiCl_4$ is easily hydrolysed
- Silicon tetrachloride, $SiCl_4$, is indeed easily hydrolyzed by water.
- This is due to silicon’s ability to form strong bonds with oxygen. When $SiCl_4$ comes into contact with water, it reacts to form silicic acid and hydrochloric acid.
$GeX_4$ (X = F, Cl, Br, I) is more stable than $GeX_2$
- Germanium tetrahalides $GeX_4$ are more stable than germanium dihalides $GeX_2$.
- This increased stability is due to germanium being in a higher oxidation state (+4) compared to the lower oxidation state (+2) in $GeX_2$, thereby providing greater stability.
$SnF_4$ is ionic in nature
- Tin(IV) fluoride, $SnF_4$, has a high degree of ionic character due to the presence of highly electronegative fluorine atoms and the +4 oxidation state of tin, leading to a weakened bond covalency.
- Thus, this statement is correct as $SnF_4$ is largely ionic in nature.

After evaluating all options, the statement that is incorrect is: $PbF_4$ is covalent in nature.

Which of the following is incorrect statement?

The Correct Option is A

Solution and Explanation

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