Answer:
CN^- is a strong field ligand
Explanation:
The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).
Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.
Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.
Answer:
Alkali metals
Explanation:
The alkali metals are a group of metallic elements which are present in the first group of the periodic table. In other words, they are present in group 1 of the periodic table. These elements have one electron in their valence shell, the reason why they are placed in group 1.
They ionize by losing one electron to achieve the configuration of the nearest Noble gas or inert gas. Because they need to offset only one electron in their outermost shell, they are very chemically reactive and hence rarely occur in the free state.
Examples of elements in this group include lithium, potassium and sodium. They each have one electron only in their outermost shells.
Answer:
-973 KJ
Explanation:
The balanced reaction equation is;
N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)
Reduction potential of hydrazine = -1.16 V
Reduction potential of chlorine = 1.36 V
From;
E°cell= E°cathode - E°anode
E°cell= 1.36 - (-1.16)
E°cell= 2.52 V
∆G°=- nFE°cell
n= number of moles of electrons = 4
F= Faraday's constant = 96500 C
E°cell = 2.52 V
∆G°=- (4 × 96500 × 2.52)
∆G°= -972720 J
∆G°= -972.72 KJ
Answer:
D. 0.3 M
Explanation:
NH4SH (s) <--> NH3 (g) + H2S (g)
Initial concentration 0.085mol/0.25L 0 0
Change in concentration -0.2M +0.2 M +0.2M
Equilibrium 0.035mol/0.25 L=0.14M 0.2M 0.2M
concentration
Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M
K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
