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

Let's analyze the given statements one by one to determine which is incorrect:

$GeX_4$ (X= F, Cl, Br, I) is more stable than $GeX_2$:
Germanium (Ge) belongs to Group 14, which typically forms tetravalent compounds like $GeX_4$. These are generally more stable than divalent compounds like $GeX_2$. Thus, this statement is correct.
$SnF_4$ is ionic in nature:
The nature of a compound as ionic or covalent is determined by the difference in electronegativity between the atoms. Tin(IV) fluoride ($SnF_4$) is largely ionic due to the substantial electronegativity difference between Sn and F. Hence, this statement is correct.
$PbF_4$ is covalent in nature:
Lead typically forms covalent compounds in higher oxidation states due to the inert pair effect, making $PbF_4$ more covalent. Nevertheless, in reality, $PbF_4$ is more ionic because it involves a significant difference in electronegativity between Pb and F, leading to ionic character. Therefore, this statement is incorrect.
$SiCl_4$ is easily hydrolyzed:
Silicon tetrachloride ($SiCl_4$) is easily hydrolyzed in water to form Si(OH)₄ and HCl because silicon has vacant d-orbitals that can accept lone pairs from water molecules. This statement is correct.

Based on this analysis, the incorrect statement is: $PbF_4$ is covalent in nature.

Answer 2

Let's analyze the given statements one by one to determine which is incorrect:

$GeX_4$ (X= F, Cl, Br, I) is more stable than $GeX_2$:
Germanium (Ge) belongs to Group 14, which typically forms tetravalent compounds like $GeX_4$. These are generally more stable than divalent compounds like $GeX_2$. Thus, this statement is correct.
$SnF_4$ is ionic in nature:
The nature of a compound as ionic or covalent is determined by the difference in electronegativity between the atoms. Tin(IV) fluoride ($SnF_4$) is largely ionic due to the substantial electronegativity difference between Sn and F. Hence, this statement is correct.
$PbF_4$ is covalent in nature:
Lead typically forms covalent compounds in higher oxidation states due to the inert pair effect, making $PbF_4$ more covalent. Nevertheless, in reality, $PbF_4$ is more ionic because it involves a significant difference in electronegativity between Pb and F, leading to ionic character. Therefore, this statement is incorrect.
$SiCl_4$ is easily hydrolyzed:
Silicon tetrachloride ($SiCl_4$) is easily hydrolyzed in water to form Si(OH)₄ and HCl because silicon has vacant d-orbitals that can accept lone pairs from water molecules. This statement is correct.

Based on this analysis, the incorrect statement is: $PbF_4$ is covalent in nature.

Which of the following is incorrect statement ?

The Correct Option is C

Solution and Explanation

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