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Why are d orbitals sometimes used to form hybrid orbitals? Which period of elements does not use d orbitals for hybridization? If necessary, which d orbitals (3d, 4d, 5d, or 6d) would sulphur use to form hybrid orbitals requiring d atomic orbitals? Answer the same question for arsenic and for iodine.

Short Answer

Expert verified

d-Orbitals are used for the excess electrons. The second period electrons will not use the d- orbitalsSulphur uses 3d orbitals, arsenic uses 4dand iodine uses 5d orbitals.

Step by step solution

01

Reason for using d-orbitals in hybrid orbitals

When the central atom is not following the octet rule, we use d-orbitals. The central atom has more than 8 electrons around it. The d orbitals in hybridization are then used for the arrangement of the excess electrons.

The second period electrons will not use the d- orbitals since they would have not excess electrons in the valence orbital and the second d- orbital does not exist because of higher energy.

02

d-orbital used in different elements

Sulphur is included in the third period if the central atom contains more than 8 electrons, then it will use the 3d- orbital to be able to undergo hybridization.

For Arsenic, it will use the 4d- orbitals for hybridization.

For iodineit will use the 5d orbitals for hybridization.

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Most popular questions from this chapter

Use the localized electron model to describe the bonding in CCl4.

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions:

NCl3(g)โ†’NCl2(g)+Cl(g)ฮ”H=375kJ/molONCl(g)โ†’NO(g)+Cl(g)ฮ”H=158kJ/mol
Rationalize the difference in the values ofฮ”Hfor these reactions, even though each reaction appears to involve the breaking of only one N-Cl bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)

The sp2hybrid atomic orbitals have the following general form:

role="math" localid="1663746599823" ฯ•1=13ฯ•s+2Aฯ•pxฯ•2=13ฯ•s-2Aฯ•px+Bฯ•pyฯ•3=13ฯ•s-2Aฯ•px-Bฯ•py

where, ฮฆs,ฮฆpxandrole="math" localid="1663746751426" ฮฆpyrepresent orthonormal (normalized and orthogonalized) atomic orbitals. Calculate the values of Aand B.

Consider three molecules: A, B, and C. Molecule A has a hybridization of sp3. Molecule B has two more effective pairs (electron pairs around the central atom) than molecule A. Molecule C consists of two s bonds and two p bonds. Give the molecular structure, hybridization, bond angles, and an example for each molecule.

For each of the following chemical formulas, an NMR spectrum is described, including relative overall areas (intensities) for the various signals given in parentheses. Draw the structure of a compound having the specific formula that would give the described NMR spectrum. Hint: All of these formulas represent organic compounds. Lewis structures for organic compounds typically have all atoms in the compound with a formal charge of zero. This is the case in this problem.)

a.C2H3Cl3;NMRhas one singlet signal.

b.C3H6Cl2;NMRhas a triplet (4) and a quintet (2) signal.

c.C3H6O2; NMR has a singlet (1), a quartet (2), and a triplet (3)signal.

d.C5H10O; NMR has a heptet (1), a singlet (3), and a doublet (6) signal.

e.C3H6O;NMRhas a triplet (3), a quintet (2), and a triplet (1) signal.

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