Question

Give the Hybridisation in SF4 and XeF4 molecule.​

Answer 1

Explanation:

Answer:

\boxed{\mathfrak{Work \ done \ by \ the \ force = zero}}Work done by the force=zero

Given:

Force \rm (\overrightarrow{F}) = 4 \hat{i} + 5\hat{j}(F)=4i^+5j^

Displacement \rm (\overrightarrow{d}) = 5 \hat{i} - 4 \hat{j}(d)=5i^−4j^

Explanation:

Work done (W) is dot product of force vector and displacement vector i.e.

\boxed{ \bold{ W = \overrightarrow{F}.\overrightarrow{d}}}W=F.d

By substituting values in the equation we get:

\begin{gathered}\rm \implies W = (4\hat{i} + 5\hat{j} ).(5\hat{i} - 4\hat{j} ) \\ \\ \rm \implies W =4 \times 5 - 5 \times 4 \\ \\ \rm \implies W = 20 - 20 \\ \\ \rm \implies W = 0 \: J\end{gathered}⟹W=(4i^+5j^).(5i^−4j^)⟹W=4×5−5×4⟹W=20−20⟹W=0J

$Answer:</p><p>\boxed{\mathfrak{Work \ done \ by \ the \ force = zero}} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ Work done by the force=zero</p><p>Given:</p><p>Force \rm (\overrightarrow{F}) = 4 \hat{i} + 5\hat{j}(F)=4i^+5j^</p><p>Displacement \rm (\overrightarrow{d}) = 5 \hat{i} - 4 \hat{j}(d)=5i^−4j^</p><p>Explanation:</p><p>Work done (W) is dot product of force vector and displacement vector i.e.</p><p>\boxed{ \bold{ W = \overrightarrow{F}.\overrightarrow{d}}}W=F.d</p><p>By substituting values in the equation we get:</p><p>\begin{gathered}\rm \implies W = (4\hat{i} + 5\hat{j} ).(5\hat{i} - 4\hat{j} ) \\ \\ \rm \implies W =4 \times 5 - 5 \times 4 \\ \\ \rm \implies W = 20 - 20 \\ \\ \rm \implies W = 0 \: J\end{gathered}⟹W=(4i^+5j^).(5i^−4j^)⟹W=4×5−5×4⟹W=20−20⟹W=0J</p><p>$