It's a combination of factors: Less electrons paired in the same orbital More electrons with parallel spins in separate orbitals Pertinent valence orbitals NOT close enough in energy for electron pairing to be stabilized enough by large orbital size DISCLAIMER: Long answer, but it's a complicated issue, so... :) A lot of people want to say that it's because a "half-filled subshell" increases stability, which is a reason, but not necessarily the only reason. However, for chromium, it's the significant reason. It's also worth mentioning that these reasons are after-the-fact; chromium doesn't know the reasons we come up with; the reasons just have to be, well, reasonable. The reasons I can think of are: Minimization of coulombic repulsion energy Maximization of exchange energy Lack of significant reduction of pairing energy overall in comparison to an atom with larger occupied orbitals COULOMBIC REPULSION ENERGY Coulombic repulsion energy is the increased energy due to opposite-spin electron pairing, in a context where there are only two electrons of nearly-degenerate energies. So, for example... ↑ ↓ −−−−−
↑ ↓ −−−−−
↑ ↓ −−−−− is higher in energy than ↑ ↓ −−−−−
↓ ↑ −−−−−
↑ ↓ −−−−− To make it easier on us, we can crudely "measure" the repulsion energy with the symbol Π c . We'd just say that for every electron pair in the same orbital, it adds one Π c unit of destabilization. When you have something like this with parallel electron spins... ↑ ↓ −−−−−
↑ ↓ −−−−−
↑ ↓ −−−−− It becomes important to incorporate the exchange energy. EXCHANGE ENERGY Exchange energy is the reduction in energy due to the number of parallel-spin electron pairs in different orbitals. It's a quantum mechanical argument where the parallel-spin electrons can exchange with each other due to their indistinguishability (you can't tell for sure if it's electron 1 that's in orbital 1, or electron 2 that's in orbital 1, etc), reducing the energy of the configuration. For example... ↑ ↓ −−−−−
↑ ↓ −−−−−
↑ ↓ −−−−− is lower in energy than ↑ ↓ −−−−−
↓ ↑ −−−−−
↑ ↓ −−−−− To make it easier for us, a crude way to "measure" exchange energy is to say that it's equal to Π e for each pair that can exchange. So for the first configuration above, it would be stabilized by Π e ( 1 ↔ 2 ), but the second configuration would have a 0 Π e stabilization (opposite spins; can't exchange). PAIRING ENERGY Pairing energy is just the combination of both the repulsion and exchange energy. We call it Π , so: Π = Π c + Π e
Inorganic Chemistry, Miessler et al. Inorganic Chemistry, Miessler et al. Basically, the pairing energy is: higher when repulsion energy is high (i.e. many electrons paired), meaning pairing is unfavorable lower when exchange energy is high (i.e. many electrons parallel and unpaired), meaning pairing is favorable So, when it comes to putting it together for chromium... ( 4 s and 3 d orbitals) ↑ ↓ −−−−− ↑ ↓ −−−−−
↑ ↓ −−−−−
↑ ↓ −−−−−
↑ ↓ −−−−−
↑ ↓ −−−−− compared to ↑ ↓ −−−−− ↑ ↓ −−−−−
↑ ↓ −−−−−
↑ ↓ −−−−−
↑ ↓ −−−−−
↑ ↓ −−−−− is more stable. For simplicity, if we assume the 4 s and 3 d electrons aren't close enough in energy to be considered "nearly-degenerate": The first configuration has Π = 10 Π e . (Exchanges: 1 ↔ 2 , 1 ↔ 3 , 1 ↔ 4 , 1 ↔ 5 , 2 ↔ 3 ,
2 ↔ 4 , 2 ↔ 5 , 3 ↔ 4 , 3 ↔ 5 , 4 ↔ 5 ) The second configuration has Π = Π c + 6 Π e . (Exchanges: 1 ↔ 2 , 1 ↔ 3 , 1 ↔ 4 , 2 ↔ 3 , 2 ↔ 4 , 3 ↔ 4 ) Technically, they are about 3.29 eV apart (Appendix B.9), which means it takes about 3.29 V to transfer a single electron from the 3 d up to the 4 s . We could also say that since the 3 d orbitals are lower in energy, transferring one electron to a lower-energy orbital is helpful anyways from a less quantitative perspective. COMPLICATIONS DUE TO ORBITAL SIZE Note that for example, W has a configuration of [ X e ] 5 d 4 6 s 2 , which seems to contradict the reasoning we had for Cr , since the pairing occurred in the higher-energy orbital. But, we should also recognize that 5 d orbitals are larger than 3 d orbitals, which means the electron density can be more spread out for W than for Cr , thus reducing the pairing energy Π . That is, Π W
the electron configuration of 1s^2 2s^2 2p^6 3s^2 3p^6, can also be written in noble gas configuration as [Ne] 3s^2 3p^6 which translates to the element Argon ( Ar ). Since the outer shell is full with 8 valence electrons, there is no lone electron so all elections are paired.
7.5 mol of hydrogen would be needed to consume the available nitrogen.
Explanation:
When hydrogen reacts with nitrogen, ammonia is formed as shown below;
3H₂ (g) + N₂ (g) → 2NH₃ (g)
As seen from the equation, every 3 moles of H₂ react with a mole of N₂ to form 2 moles of NH₃.
The limiting factor in a chemical reaction is the reactant that gets depleted first.
Because the molar mass of nitrogen gas is approximately 28g/mol, 70g of nitrogen gas would be 2.5 moles.
The reaction ratio of nitrogen to hydrogen in the reaction is 1 : 3. The reaction would require 2.5 * 3 (7.5) moles of hydrogen for a complete reaction.
However since there are only 7g on hydrogen, (Remember 1 mole of H₂ is approximately 2g), the available moles of H₂ is 7 / 2 = 3.5
3.5 moles fall short of the 7.5 moles of H₂ required for a complete reaction. H₂ gets depleted first before N₂. The reaction would require 4 more moles of H₂.
1. They are produced by the hottest and most energetic objects in the universe, such as neutron stars and pulsars, supernova explosions, and regions around black holes.
2. On Earth, gamma waves are generated by nuclear explosions, lightning, and the less dramatic activity of radioactive decay.
3. Scientists can use gamma rays to determine the elements on other planets.
