如果你也在 怎样代写无机化学inorganic chemistry这个学科遇到相关的难题,请随时右上角联系我们的24/7代写客服。无机化学inorganic chemistry涉及到无机和有机金属化合物的合成和行为。这个领域涵盖了非碳基的化合物,这些化合物是有机化学的主题。这两门学科之间的区别远非绝对,因为有机金属化学的分支学科有很多重叠。它在化学工业的各个方面都有应用,包括催化、材料科学、颜料、表面活性剂、涂料、药物、燃料和农业。
无机化学inorganic chemistry许多无机化合物是离子化合物,由阳离子和阴离子通过离子键连接组成。盐(属于离子化合物)的例子有氯化镁MgCl2,它由镁的阳离子Mg2+和氯的阴离子Cl-组成;或氧化钠Na2O,它由钠的阳离子Na+和氧化阴离子O2-组成。在任何盐中,离子的比例是这样的:电荷相互抵消,因此大部分化合物是电中性的。离子由其氧化状态描述,其形成的难易程度可以从电离电位(阳离子)或从母元素的电子亲和力(阴离子)推断出来。
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化学代写|无机化学作业代写inorganic chemistry代考|Spatial arrangement of electron pairs around the central atom of a given molecule/ion
The electron already present in the valence shell of the central atom of the given species plus the electron acquired by the central atom as a result of bonding with other atoms are called the valence shell electrons. Half of this gives the number of electron pairspresent in the valence shell of the central atom of the given species.
Electron pairs present in the valence shell of the central atom occupy localized orbitals which arrange themselves in space in such a way that they keep apart from one another as for as possible. This gives minimum energy and maximum stability to the species. As there can be only one definite orientation of orbitals corresponding to minimum energy, a molecule or ion of a given substance has a definite shape, that is, a definite geometry.
When the central atom in a molecule is surrounded by bonded electron pair only, the molecule will have a regular geometry or shape. The geometry depends on the number of bonded electron pairs as given in Table 1.1.
化学代写|无机化学作业代写inorganic chemistry代考|Regular and irregular geometry: presence of hybrid orbitals containing bond pairs and lone pairs
If the central atom is surrounded only by orbitals containing shared pair of electrons, that is, bps), and there are no hybrid orbitals containing lone pairs (lps) of electrons in the valence shells, the molecule has a regular geometry.
If, however, the central atom is surrounded by one or more hybrid orbitals containing lone pairs of electrons in the valence shells, the bond angle gets distorted from the value expected for a particular geometry of molecule or ions. Hence, with the change in the magnitude of the bond angles, the shape of the molecule or ions gets distorted.
Cause of distortion in geometry/change in bond angles
The strength of the repulsions between the electron pairs in a valence shell decrease in the order: lone pair (lp)-lone pair $(\mathrm{lp})>$ lone pair $(\mathrm{lp})$-bond pair $(\mathrm{bp})>$ bond pair (bp)-bond pair (bp). In other words, lone pairs and bond pairs are not equivalent but lone pairs exert a greater effect on bond angles than bond pairs.
This rule can be easily understood because a bonding electron pair is under the influence of two nuclei (two +ve centres) whereas a lone pair is under the influence of only one nucleus. Thus, a lone pair is expected to occupy a broader orbital with a greater electron density radially distributed closer to the central atom than the bond pair electrons, which are drawn out between two positive centres. In other words, the lone pair occupies more space on the surface of the central atom than a bonding pair. Hence, it will repel the electron pair in the neighbouring orbitals more strongly. This is shown diagrammatically in Figure 1.1.
Consequently, the presence of one or more orbitals with lone pairs has the effect of altering the bond angles to a significant extent. The molecule will not retain any regular geometry now.
化学代写|无机化学作业代写inorganic chemistry代考|Effect of electronegativity: repulsions exerted by bond pairsdecrease as the electronegativity of the bonded atom increases
The high electronegativity of the bonded atom (with the central atom) pulls the $\sigma$-bonded electron pair away from the central atom nucleus, thereby contracting and thinning out the orbital.
Hence, more the electronegative of the bonded atom, more the electron density is displaced towards it in the $\sigma$-bond. This reduces repulsion between the bond pairs, and hence allows the lone pair(s) to expand more.
Both the factors causes decrease in the bond angles. The reduction in bond angle with increase in electronegativity may be viewed in the following examples:
$$
\begin{aligned}
&\mathrm{NH}{3}\left(107.3^{\circ}\right)>\mathrm{NF}{3}\left(102^{\circ}\right) \
&\mathrm{OH}{2}\left(104.45^{\circ}\right)>\mathrm{OF}{2}\left(103.2^{\circ}\right) \
&\mathrm{PI}{3}\left(102^{\circ}\right)>\mathrm{PBr}{3}\left(101.5^{\circ}\right)>\mathrm{PCl}{3}\left(100^{\circ}\right) \ &\operatorname{AsI}{3}\left(101^{\circ}\right)>\operatorname{AsBr}{3}\left(100.5^{\circ}\right)>\operatorname{AsCl}{3}\left(98.4^{\circ}\right)
\end{aligned}
$$
无机化学代写
化学代写|无机化学作业代写INORGANIC CHEMISTRY代考|SPATIAL ARRANGEMENT OF ELECTRON PAIRS AROUND THE CENTRAL ATOM OF A GIVEN MOLECULE/ION
已经存在于给定物质的中心原子的价壳层中的电子加上中心原子由于与其他原子键合而获得的电子称为价壳层电子。其中一半给出了给定物质中心原子的价壳中存在的电子对数。
存在于中心原子的价壳中的电子对占据局部轨道,这些轨道在空间中以这样一种方式排列,即它们尽可能地彼此分开。这为物种提供了最小的能量和最大的稳定性。由于与最小能量对应的轨道只能有一个确定的方向,因此给定物质的分子或离子具有确定的形状,即确定的几何形状。
当分子中的中心原子仅被键合电子对包围时,分子将具有规则的几何形状或形状。几何形状取决于表 1.1 中给出的键合电子对的数量。
化学代写|无机化学作业代写INORGANIC CHEMISTRY代考|REGULAR AND IRREGULAR GEOMETRY: PRESENCE OF HYBRID ORBITALS CONTAINING BOND PAIRS AND LONE PAIRS
如果中心原子仅被包含共享电子对的轨道(即bps)包围,并且不存在包含孤对电子的混合轨道一世ps价壳层中的电子,分子具有规则的几何形状。
然而,如果中心原子被一个或多个在价壳层中包含孤对电子的杂化轨道包围,则键角会与特定几何形状的分子或离子的预期值不同。因此,随着键角大小的变化,分子或离子的形状会发生扭曲。
几何变形/键角变化的原因
价壳中电子对之间的排斥强度按顺序降低:孤对电子一世p-孤独的一对(一世p)>单对(一世p)-键对(bp)>键对bp-键对bp. 换句话说,孤对和键对不等价,但孤对对键角的影响比键对更大。
这个规则很容易理解,因为一个键合电子对受到两个原子核的影响吨在这+v和C和n吨r和s而孤对只受一个核的影响。因此,与在两个正中心之间拉出的键对电子相比,预计孤对电子占据更宽的轨道,径向分布更靠近中心原子的电子密度更大。换句话说,孤对在中心原子表面上比键对占据更多的空间。因此,它将更强烈地排斥相邻轨道中的电子对。这在图 1.1 中以图解方式显示。
因此,一个或多个孤对轨道的存在会在很大程度上改变键角。该分子现在不会保留任何规则的几何形状。
化学代写|无机化学作业代写INORGANIC CHEMISTRY代考|EFFECT OF ELECTRONEGATIVITY: REPULSIONS EXERTED BY BOND PAIRSDECREASE AS THE ELECTRONEGATIVITY OF THE BONDED ATOM INCREASES
键合原子的高电负性在一世吨H吨H和C和n吨r一种一世一种吨这米拉σ键合电子对远离中心原子核,从而使轨道收缩和变薄。
因此,键合原子的电负性越大,在σ-键。这减少了键对之间的排斥,因此允许孤对s扩大更多。
这两个因素都会导致键角减小。键角随着电负性的增加而减小,可以从以下例子中看出:
$$
\begin{aligned}
&\mathrm{NH}{3}\left(107.3^{\circ}\right)>\mathrm{NF}{3}\left(102^{\circ}\right) \
&\mathrm{OH}{2}\left(104.45^{\circ}\right)>\mathrm{OF}{2}\left(103.2^{\circ}\right) \
&\mathrm{PI}{3}\left(102^{\circ}\right)>\mathrm{PBr}{3}\left(101.5^{\circ}\right)>\mathrm{PCl}{3}\left(100^{\circ}\right) \ &\operatorname{AsI}{3}\left(101^{\circ}\right)>\operatorname{AsBr}{3}\left(100.5^{\circ}\right)>\operatorname{AsCl}{3}\left(98.4^{\circ}\right)
\end{aligned}
$$
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