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28%
Explanation:
mass of solute(KBr) = 3.73g
mass of solvent(H2O) = 131g
mass of solution = mass of solute + mass of solvent
= 3.73 + 131
= 134.73g
![\sf \large {mass \: percentage = \frac{mass \: of \: solute}{mass \: of \: solvent} \times 100} \\ \\ \sf mass \: percentage = \frac{3.73}{134.73} \times 100 \\ \\ \sf mass \: percentage = 0.028 \times 100 \\ \\ \sf mass \: percentage = 28\%](https://tex.z-dn.net/?f=%20%5Csf%20%5Clarge%20%7Bmass%20%5C%3A%20percentage%20%3D%20%20%5Cfrac%7Bmass%20%5C%3A%20of%20%5C%3A%20solute%7D%7Bmass%20%5C%3A%20of%20%5C%3A%20solvent%7D%20%20%5Ctimes%20100%7D%20%5C%5C%20%20%5C%5C%20%20%5Csf%20%20mass%20%5C%3A%20percentage%20%3D%20%20%5Cfrac%7B3.73%7D%7B134.73%7D%20%20%5Ctimes%20100%20%5C%5C%20%20%5C%5C%20%20%5Csf%20mass%20%5C%3A%20percentage%20%3D%20%200.028%20%5Ctimes%20100%20%5C%5C%20%20%5C%5C%20%20%5Csf%20mass%20%5C%3A%20percentage%20%3D%2028%5C%25)
Fe2O3 + 3C → 2Fe + 3CO :)
<span>Xe = VIII = 8 valence electrons
F = VII = 4 (7 ve) = 28 valence electrons</span>
total ve = 8 + 28 = 36 ve
<span>36 - 4(2) = 28 ve
(there are 2 electrons in each bond x 4 bonds)</span>
<span>28 - 4(6) = 4
(We assign the remaining electrons to F atoms)</span>
<span>4 - 2(2) = 0
(Therefore 4 electrons left => we have 2 lone pairs)</span>
The steric number = No. of
σ bonds + #lone pairs
= 4 σ bonds + 2 lone pairs
= 6 => d²sp³ (6 hybrid orbitals)
<span>4 bonds + 2 lone pairs
=> square planar</span>