SF6 has six bonding pairs and no lone pairs, resulting in an octahedral electron pair geometry.

Chlorine ion (Cl-) has gained an electron, resulting in the configuration 1s2 2s2 2p6.

Oxygen has 8 electrons. The electronic configuration is 1s2 2s2 2p4.

Sodium has 11 electrons, and its configuration is 1s2 2s2 2p6 3s1.

Sodium ion (Na+) has lost one electron, resulting in the configuration 1s2 2s2 2p6.

Chlorine (Cl) has 17 electrons, so its configuration is 1s2 2s2 2p6 3s2 3p5.

Iron (Fe) has the configuration [Ar] 4s2 3d6.

Oxygen has 8 electrons, and its electronic configuration is 1s2 2s2 2p4.

The electronic configuration of potassium (K) is 1s² 2s² 2p⁶ 3s¹.

Chloride ion has gained one electron, making its configuration 1s2 2s2 2p6 3s2 3p6.

The electronic configuration of Rutherfordium (atomic number 104) is [Rn] 5f14 6d4.

The element with atomic number 12 is magnesium (Mg), and its electronic configuration is [Ne] 3s2.

The element with atomic number 15 is phosphorus (P), and its electronic configuration is [Ne] 3s2 3p3.

Chlorine (Cl) has the configuration 1s2 2s2 2p6 3s2 3p5.

The element with atomic number 20 is Calcium, which has the configuration [Ar] 4s2.

The element with atomic number 26 is iron (Fe), and its electronic configuration is [Ar] 4s2 3d6.

Copper (Cu) has the configuration [Ar] 4s2 3d9.

The electronic configuration of Zinc (atomic number 30) is [Ar] 4s2 3d10.

The electronic configuration of Lanthanum (atomic number 57) is [Xe] 6s2.

Iron (Fe) has 26 electrons, and its electronic configuration is [Ar] 4s2 3d6.

Nitrogen has 7 electrons, and its electronic configuration is 1s2 2s2 2p3.

The potassium ion (K+) loses one electron, resulting in the configuration 1s2 2s2 2p6.

Sodium ion (Na+) has lost one electron, resulting in the configuration 1s2 2s2 2p6.

The electrostatic potential V at a distance r from a point charge Q is given by V = kQ/r, where k is Coulomb's constant.

Convert percentages to grams (assume 100 g total): 40 g C, 6.7 g H, 53.3 g O. Convert to moles: C=3.33, H=6.67, O=3.33. The simplest ratio is 1:2:1, giving CH2O.

Converting percentages to moles gives C: 40/12 = 3.33, H: 6.7/1 = 6.7, O: 53.3/16 = 3.33. The simplest ratio is 1:2:1, giving the empirical formula CH2O.

The empirical formula is obtained by dividing the subscripts by the greatest common divisor, which is 2. Thus, C6H12 becomes C3H6.

The empirical formula is obtained by dividing the subscripts by the greatest common divisor, which is 2. Thus, C6H12 becomes C3H6.

The empirical formula is the simplest whole number ratio, which is CH2O.

The empirical formula is the simplest whole number ratio of the elements in a compound. For glucose, it is CH2O.