Predict the geometry of the following species: (a) \(\mathrm{KrF}_{2}\) (b) \(\mathrm{NH}_{2} \mathrm{Cl}\) (c) \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) (d) SCN

First, we need to determine the number of valence electrons and electron groups around the central atom Kr. Kr has 7 valence electrons, and each F atom contributes 1, giving a total of 9 valence electrons. As there are 2 atoms bonded to Kr, this leaves 7 - 2 = 5 non-bonding electrons or 5/2 = 2.5 non-bonding pairs (recall 5 because it's from the central atom Kr). 2.5 pairs are not possible; thus, we understand that Kr can expand its octet and can accommodate more than 8 electrons in its valence shell. So, Kr utilizes the d-orbitals to create 5 electron groups around it. This leads to a trigonal bipyramidal arrangement, with 2 F atoms and 3 lone pairs. As lone pairs occupy the equatorial positions for least repulsion, KrF2 will have a linear shape.

First, determine the number of valence electrons and electron groups around the central atom N. N has 5 valence electrons, and there are 6 electrons coming from the H and Cl atoms, giving a total of 11 valence electrons. 3 atoms are bonded to N, so there are 11 - 3 = 8 non-bonding electrons, or 8/2 = 4 non-bonding pairs. These pairs combine with the 3 atoms bonded to N to form 4 electron groups around it. This leads to a tetrahedral electron pair geometry. Since there are 3 atoms and 1 lone pair, the molecular geometry of NH2Cl is trigonal pyramidal.

First, determine the number of valence electrons and electron groups around the central atom C. C has 4 valence electrons, and there are 8 electrons from the H and Br atoms, giving a total of 12 valence electrons. Since all 4 atoms are bonded to C, there are no non-bonding electrons. This leads to 4 electron groups around C, giving a tetrahedral electron pair geometry. As there are no lone pairs on the central atom, the molecular geometry of CH2Br2 is also tetrahedral.

Determine the number of valence electrons and electron groups around the central atom C. C has 4 valence electrons, and there are 6 electrons coming from the N and S atoms, giving a total of 10 valence electrons. As SCN− has a charge of -1, we add another electron, giving us 11 valence electrons around the central atom. 2 atoms are bonded to C, so there are 11 - 2 = 9 non-bonding electrons, or 9/2 = 4.5 non-bonding pairs. However, 4.5 pairs are not possible, so we see that there is a double bond between the N and C atoms. This reduces the non-bonding electrons to 7, which gives 3.5 non-bonding pairs. Once again, an impossible scenario. Finally, we conclude that there's a triple bond between the C and N atoms, and a single bond between the C and S atoms. This gives us 2 electron groups around the central atom. Hence, the molecular geometry of SCN− is linear.