<\/p>\n

Yeni Mekanikler, Fizikler ve Grafikler<\/h2>\n

Lewis yap\u0131s\u0131 \u2014 kimyada bir molek\u00fcldeki elektron da\u011f\u0131l\u0131m\u0131n\u0131 g\u00f6stermek i\u00e7in kullan\u0131lan grafiksel bir g\u00f6sterimdir. Bu kavram\u0131n anla\u015f\u0131lmas\u0131yla \u2014 farkl\u0131 kimyasal elementlerin \u00f6zelliklerini ve davran\u0131\u015flar\u0131n\u0131 daha iyi anlamak m\u00fcmk\u00fcn olur. \u00d6rne\u011fin, oksijen atomunun 6 de\u011ferlik elektronu vard\u0131r. Yap\u0131s\u0131n\u0131 temsil etmek i\u00e7in demir kama Bir atomun boyutunu belirlemek i\u00e7in \u00f6ncelikle de\u011ferlik elektronlar\u0131n\u0131n say\u0131s\u0131n\u0131 saymam\u0131z gerekir.<\/p>\n

De\u011ferlik elektronlar , atomun son y\u00f6r\u00fcngesinde bulunan elektron say\u0131s\u0131, elementin sembol\u00fc etraf\u0131nda noktalar ile g\u00f6sterilir. Bu \u015fekilde \u2014 Cr i\u00e7in g\u00f6r\u00fcnt\u00fcn\u00fcn Lewis yap\u0131s\u0131 2 O 7 2- ula\u015f\u0131l\u0131r. Bulunan ayn\u0131 nedenlerden dolay\u0131, permanganat iyonunda negatif formal y\u00fckl\u00fc iki oksijen atomu b\u0131rak\u0131lmal\u0131d\u0131r; bunlar toplamda dikromat iyonunun y\u00fck\u00fc olan -2’ye e\u015fit olur.<\/p>\n

Dolay\u0131s\u0131yla (karbonda d\u00f6rtten fazla ba\u011f veya hidrojende birden fazla ba\u011f varsa), taslak at\u0131l\u0131p yeni ve daha ger\u00e7ek\u00e7i bir taslak olu\u015fturulabilir. K\u0131sacas\u0131 Lewis yap\u0131s\u0131 Atomlar aras\u0131ndaki kimyasal ba\u011flar\u0131 anlamak i\u00e7in \u00f6nemli bir ara\u00e7t\u0131r, \u00e7\u00fcnk\u00fc de\u011ferlik kabu\u011fundaki elektron da\u011f\u0131l\u0131m\u0131n\u0131 net bir \u015fekilde g\u00f6rselle\u015ftirmenizi sa\u011flar. Oksijenin 6 de\u011ferlik elektronu varken, hidrojenin 1 de\u011ferlik elektronu vard\u0131r. Lewis yap\u0131s\u0131n\u0131n basit bir \u00f6rne\u011fi, oksijenin iki hidrojen atomuna kovalent ba\u011flarla ba\u011fland\u0131\u011f\u0131 su molek\u00fcl\u00fcd\u00fcr (H2O). Atomlar etraf\u0131ndaki elektron da\u011f\u0131l\u0131m\u0131n\u0131 anlamak ve molek\u00fcler geometriyi tahmin etmek i\u00e7in kullan\u0131l\u0131r. Bu s\u00fcre\u00e7te oyunda bulunan hatalar giderildi ve bir\u00e7ok yeni \u00f6zellik test edildi.<\/p>\n

\u00d6rne\u011fin, C bile\u015fi\u011finiz varsa 14 O 2 N 3 Karbon, oksijen ve azot i\u00e7eren gruplar\u0131 aramal\u0131s\u0131n\u0131z. Bu nedenle Lewis yap\u0131lar\u0131 her zaman dikkate al\u0131n\u0131r ve yeni kimyasal \u00f6\u011frenmeleri yo\u011funla\u015ft\u0131rabildi\u011fi i\u00e7in olduk\u00e7a faydal\u0131d\u0131r. Ancak bu \u00e7er\u00e7eve, bir atomun ve \u00e7evresinin molek\u00fcler geometrisi , kare, \u00fc\u00e7gen d\u00fczlem, \u00e7ift piramit vb. olup olmad\u0131\u011f\u0131, gibi baz\u0131 \u00f6nemli ayr\u0131nt\u0131lar\u0131 \u00f6ng\u00f6rmede ba\u015far\u0131s\u0131z olmaktad\u0131r.<\/p>\n

\u00d6nemli Notlar<\/h2>\n

<\/p>\n

Dolay\u0131s\u0131yla (hidrojen tek bir elektrona ve doldurulmaya haz\u0131r tek bir orbitale sahip oldu\u011fundan), yaln\u0131zca bir kovalent ba\u011f olu\u015fturur. Peki, \u00e7izilen \u201cmolek\u00fcler yap\u0131\u201d yerine neden C \u2013 H \u2013 H \u2013 C olmas\u0131n? Hem ba\u011f olu\u015fumunda rol oynayan hem de payla\u015f\u0131lmayan elektronlara , Br’nin hemen \u00fczerindeki yaln\u0131z elektron \u00e7ifti, kar\u015f\u0131l\u0131k gelen siyah noktalar\u0131 g\u00f6rebilirsiniz. Yukar\u0131daki g\u00f6rsel, bir Lewis yap\u0131s\u0131 \u00f6rne\u011fini g\u00f6stermektedir. Hangisinin do\u011fru oldu\u011fu, atomlar\u0131n bi\u00e7imsel y\u00fcklerine ve kimyasal yap\u0131lar\u0131na ba\u011fl\u0131 olacakt\u0131r. Ayr\u0131ca (atomlar etraf\u0131ndaki elektron da\u011f\u0131l\u0131m\u0131yla belirlenen molek\u00fcler geometri), erime ve kaynama noktalar\u0131 gibi maddenin \u00f6zelliklerini do\u011frudan etkiler.<\/p>\n

D\u00f6rt tane oldu\u011fundan \u2014 anyon i\u00e7in net y\u00fck -4 olurdu ki bu a\u00e7\u0131k\u00e7a do\u011fru de\u011fildir. Ayr\u0131ca, MnO iyonu i\u00e7in 4 – Oksijen atomlar\u0131n\u0131n formal y\u00fcklerini azaltmak i\u00e7in negatif y\u00fck\u00fcn a\u00e7\u0131\u011fa \u00e7\u0131kar\u0131lmas\u0131 gerekir. \u0130kisi de ayn\u0131d\u0131r, \u00e7\u00fcnk\u00fc yap\u0131lar ayn\u0131 molek\u00fcl form\u00fcl\u00fc C’nin yap\u0131sal izomerleri olarak ortaya \u00e7\u0131km\u0131\u015ft\u0131r. Form\u00fcl\u00fc uygulamadan \u00f6nce (hidrojenlerin tek), oksijenin iki ve karbonun d\u00f6rt ba\u011f olu\u015fturdu\u011funu ve yap\u0131n\u0131n m\u00fcmk\u00fcn oldu\u011funca simetrik olmas\u0131 gerekti\u011fini unutmay\u0131n.<\/p>\n

TEV ve TRPEV bilgisi uyguland\u0131\u011f\u0131nda, geometrinin azot yaln\u0131z \u00e7ifti taraf\u0131ndan tetrahedral olarak bozuldu\u011fu ve dolay\u0131s\u0131yla hibridizasyonunun sp oldu\u011fu sonucu \u00e7\u0131kar. Mevcut 8 elektrondan 6’s\u0131 ba\u011flara kat\u0131ld\u0131\u011f\u0131ndan, azot atomunun \u00fczerinde payla\u015f\u0131lmam\u0131\u015f bir \u00e7ift bulunur. Hesaplamalar, olu\u015fmas\u0131 gereken d\u00f6rt ba\u011f\u0131n varl\u0131\u011f\u0131n\u0131 \u00f6ng\u00f6rse de, al\u00fcminyum yeterli elektrona sahip de\u011fildir ve d\u00f6rd\u00fcnc\u00fc bir flor atomu da yoktur.<\/p>\n","protected":false},"excerpt":{"rendered":"

Bir molek\u00fcl veya iyondaki de\u011ferlik elektronlar\u0131n\u0131n ve kovalent ba\u011flar\u0131n temsili yap\u0131s\u0131d\u0131r ve molek\u00fcler yap\u0131s\u0131 hakk\u0131nda fikir edinmeye yarar. \u00d6te yandan, karbon atomunun elektron dizilimi d\u00f6rt kovalent ba\u011f\u0131n olu\u015fmas\u0131na izin verir , ve gerektirir,. Dolay\u0131s\u0131yla, asla iki ba\u011f olu\u015fturamaz , hidrojen ba\u011flar\u0131yla kar\u0131\u015ft\u0131r\u0131lmamal\u0131d\u0131r,. \u00c7\u00fcnk\u00fc her iki S-O ba\u011f\u0131n\u0131n da ayn\u0131 uzunlukta oldu\u011fu bulunmu\u015ftur. Di\u011fer taraftan deneysel […]<\/p>\n","protected":false},"author":13,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-2325","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/www.aliawais.com\/index.php?rest_route=\/wp\/v2\/posts\/2325","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.aliawais.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.aliawais.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.aliawais.com\/index.php?rest_route=\/wp\/v2\/users\/13"}],"replies":[{"embeddable":true,"href":"https:\/\/www.aliawais.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2325"}],"version-history":[{"count":1,"href":"https:\/\/www.aliawais.com\/index.php?rest_route=\/wp\/v2\/posts\/2325\/revisions"}],"predecessor-version":[{"id":2326,"href":"https:\/\/www.aliawais.com\/index.php?rest_route=\/wp\/v2\/posts\/2325\/revisions\/2326"}],"wp:attachment":[{"href":"https:\/\/www.aliawais.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2325"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.aliawais.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2325"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.aliawais.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}