Which of the following compounds has more than one some; PtCl_5NH_3, Co(BH_3)_4Cl_2, Pt(Co)_2Cl_2, Rb_3[Cr(CN)_6]
To determine which of the given compounds has more than one somo, we need to understand what a somo is.
A somo (short for "Sum Over Molecular Orbitals") is a type of molecular orbital that can be occupied by an electron pair. It is commonly found in transition metal complexes and is involved in the bonding and reactivity of these compounds.
Now let's analyze the given compounds and see if they have more than one somo:
1. PtCl5NH3:
To determine the number of somos in this compound, we need to consider the presence of ligands (NH3) and the coordination number. In this case, PtCl5NH3 has six ligands (five Cl- and one NH3) and a coordination number of 6. Since there are six ligands, there will be six available somos. Therefore, PtCl5NH3 has more than one somo.
2. Co(BH3)4Cl2:
Similar to the previous compound, we need to consider the number of ligands (BH3 and Cl-) and the coordination number. Co(BH3)4Cl2 has six ligands (four BH3 and two Cl-) and a coordination number of 6. Hence, there will be six available somos. Therefore, Co(BH3)4Cl2 has more than one somo.
3. Pt(Co)2Cl2:
In this compound, Pt is coordinated with two Co atoms and two Cl- ligands. Considering the coordination number of Pt, which is 4, Pt(Co)2Cl2 has four available somos (two from Co atoms and two from Cl- ligands). Hence, Pt(Co)2Cl2 has more than one somo.
4. Rb3[Cr(CN)6]:
In this compound, [Cr(CN)6] has a Cr atom coordinated with six CN- ligands. The coordination number of Cr is 6. Therefore, there are six available somos in [Cr(CN)6]. Since Rb3[Cr(CN)6] has three [Cr(CN)6] ions, the total number of somos is 3 times six, which is 18. Hence, Rb3[Cr(CN)6] has more than one somo.
In conclusion, all the given compounds (PtCl5NH3, Co(BH3)4Cl2, Pt(Co)2Cl2, and Rb3[Cr(CN)6]) have more than one somo.